Showing posts with label Rotational Dynamics. Show all posts
Showing posts with label Rotational Dynamics. Show all posts

Rotational Dynamics Problems with Solutions Six

Rotational dynamics is the study of motion of a rotating body about its axis of rotation. All the particles of the body other than the particles along the axis of rotation will be having angular displacement. It is the angle turned by the body with respect to its initial position about its axis of rotation. It is same for all the particles of the body even though they are at different distance from the axis of rotation. Thus we do study rotational motion in terms of different set of technical terms when compared with translatory motion. We have moment of inertia that plays the same role of mass in translatory motion and we use torque here that is similar to force while studying translatory motion. Similar to linear momentum of the body in translatory motion, we use angular momentum and its conservation in rotational dynamics.

Problem 

Two rings of equal mass and thickness made up of different materials are subjected to same torque about its geometric axis.As the materials are different, their densities are different and we need to find the angular accelerations ratio and the problem is as shown in the diagram below.


Solution

We know that torque is the product of moment of inertia and angular momentum and when the torque is constant, angular acceleration is inversely proportional to the moment of inertia. It is further given in the problem that the mass of the particles are same and the mass can be expressed as the product of volume and density of  the body. Volume can be further expressed as the product of area of cross section and the thickness of the ring. Problem can be further solved and we can find the angular acceleration ratio as shown in the diagram below.


Problem

A bullet of known mass is fired upward in a direction with a known angle and known velocity. We need to measure the angular momentum of the body when it has reached its highest point and the problem is as shown in the diagram below.


Solution

Angular momentum is defined as the moment of momentum. Its magnitude can be found as the product of mass of the particle, distance of the particle with axis of rotation and the velocity of the particle at that instant. Distance is nothing but the maximum height of the projectile and the problem can be solved as shown in the diagram below.


Problem

It is give in the problem that a cylinder of mass m rolls without slipping down an inclined plane making an angle with the horizontal. We need to measure the frictional force between the cylinder and the inclined plane is 


Solution

We know the torque as the product of moment of inertia and the angular acceleration and we know the formula for the moment of inertia. We can also apply the angular acceleration in terms of linear acceleration and and the radius of the motion. As the body is moving down, frictional force acts against the motion in the upward direction along the inclined plane. We can write the net force as the difference between component of weight and frictional force and it can be solved as shown in the diagram below.


Problem

Three masses each of mass 2 kilogram are placed at the vertices of an equilateral triangle of side 10 centimeter. We need to find the moment of inertia about an axis passing through any vertex and perpendicular to the plane of the triangle. Problem is as shown in the diagram below.


Solution

Moment of inertia of the system is the sum of moment of inertia of each particle of the system and it can be measured in terms of radius of gyration also. It is the effective distance of total mass of the particle concentrated at a point from the axis of rotation.


Problem

Reel of thread in the form of a solid cylinder of mass M and radius r is allowed to unroll by holding the loose end of thread in hand. We need to find the acceleration with which the reel falls down and the problem is as shown in the diagram below.


Solution

We can write equation of motion to a body  in the terms of force as shown in the diagram below. By applying the definition of the torque also, we can solve the problem as shown in the diagram below.


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Rotational Dynamics Problems with Solutions Five

We are solving series of problems on rotational dynamics. We are solving problems based on moment of inertia here. Moment of inertia is the ability of a body to hold its position against rotational motion of a body. We shall apply toque to overcome moment of inertia of a body. Moment of inertia depends on the size, shape of the body and the axis of rotation. Moment of inertia changes with axis of rotation as it changes the distance of the particles from axis of rotation.To find moment of inertia of a body about different symmetrical axis, we have parallel and perpendicular axes theorems.

Problem

A mass m is released with a horizontal speed from the top of a smooth and fixed hemispherical bowl or radius r. We need to find the angle it makes with vertical when it leaves the contact of the surface of the bowl. Problem is as shown in the diagram below.


Solution

When the ball leaves the surface of the bowl, its normal reaction becomes zero because of lack of contact from the surface. We also can find that the component of the weight acts like a centripetal force. We need to apply law of conservation of energy at both the points so that we can solve the problem as shown in the diagram below.


Problem

Two boys of mass 10 and 8 kilogram are moving along a vertical rope where one is climbing up and other is climbing down with a constant acceleration. We need to find the tension in the rope at fixed support and the problem is as shown in the diagram as shown below.


Solution

Tension in the string becomes different at different places of the string as shown in the diagram below. We can write equations of motion for each body as shown. The resultant force on each body is the effective force acting on it. Force acting in the same direction shall be treated as positive and the force acting against the motion shall be treated as negative. We can solve the problem as shown in the diagram below.


Problem

Moment of inertia of a solid sphere about its diameter I. If that sphere is recast into eight identical small spheres, then the moment of inertia of each sphere about its diameter and the problem is as shown in the diagram below.


Solution

We know the formula of moment of inertia basing on the standard formula. Volume of big sphere is equal to total volume of small spheres. Thus we can find the relation between the radius of small and big spheres. We can find the moment of inertia of small spheres as shown in the diagram below.


Problem

The length of solid sphere is 4.5 times its radius and its moment of inertia is given as I. If this solid cylinder is recasted  into a solid sphere, we need to find moment of inertia of a solid sphere and the problem is as shown in the diagram below.


Solution

We know the formula for the moment of inertia of a solid cylinder as well as sphere. Even if the body is recasted, its mass remains constant. By applying the condition, we can solve the problem as shown in the diagram below.


Problem

Four small spheres each of radius r and mass m are placed with their centers on four corners of a square of side L. We need to find the moment of inertia of the system as shown in the diagram below.


Solution

We need to find the moment of inertia of the system and it is sum of moment of inertia of all four spheres together. We need to apply parallel axes theorem and solve the problem as shown in the diagram below.


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Rotational Dynamics Problems with Solutions Four

We are solving series of problems on the concepts of rotational dynamics. When no external force is acting on a system, the algebraic sum of moments about center of mass is zero. It means that the moments on that particle from one side of the particles is similar to moments about the other side particles. Moment is the product of distance of the particle from a reference point and mass of the particle. Angular momentum is defined as the moment of momentum. It is conserved when no external torque is acting on the system.

Problem

A point sized body of known mass is moving at a velocity of four meter per second along a straight line and it is represented by the given equation as shown in the problem. We need to find the angular momentum of the system.


Solution

The given equation is in the form a mathematical equation that represents a straight line as shown in the diagram below. Basing on that we can find the perpendicular distance between the point of action and the body. We can find angular momentum as the product of distance with the linear momentum.


Problem

A ring and disc having the same mass roll without slipping with the same linear velocity. If the kinetic energy of the ring is 8 joule, we need to measure the kinetic energy of the disc.


Solution

As the body is rolling it will have rotational kinetic energy and as it is having translatory motion, it also has translatory kinetic energy and the total energy is the sum of both of them. We can express rotational kinetic energy in terms of transnational kinetic energy as shown in the diagram below. It depends on radius of gyration of the body and it can be written for both ring and disc. Problem can be further simplified as shown in the diagram below.


Problem

Two point sized heavy masses are attached to a single mass less string at two different points and they are moved in the horizontal circle. The ratio of distances of particles from center of circles is  
 1 :2. If the tension between the particles is four newton, we need to measure the tension in the remaining string. Problem is as shown in the diagram below.


Solution

We know that when a body is under horizontal circular motion, centripetal force is nothing but the tension in the wire at respective points. We can measure it and simplified further as shown in the diagram below.


Problem

A particle goes in a horizontal circular path on a smooth inner surface of conical funnel whose vertex angle is 90 degree. If the height of the plane of the circle above the vertex is 9.8 centimeter, we need to find the speed of the particle. Problem is as shown in the diagram below.


Solution

We know that the weight of the particle always acts in vertically downward direction and its reaction does acts perpendicular to the point of contact as shown in the diagram below. We can resolve the reaction into components and SIN component of reaction compensates the weight of the particle at that given instant. The COS component of the reaction acts towards the center and it acts line centripetal force. By equating them correspondingly and further simplifying, we can get velocity as shown in the diagram below.


Problem

A particle of known mass is fixed to one end of a light spring of known force constant and length of the wire is given to us. The system is rotated about the other end of the spring with a known angular velocity in the gravity free space. We need to measure the increase in the length of the spring and the problem is as shown in the diagram below.


Solution

We can define force constant basing on the nature of the spring and we can also express the work done in terms of applied force and small extension of the spring. Problem can be further simplified as shown in the diagram below.


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Rotational Dynamics Problems with Solutions Three

We are solving series of problems on rotational dynamics. Similar to equations of motion of translatory motion, we can write equations of rotational dynamics. Similar to mass we have moment of inertia and similar to force, we have torque in rotational dynamics. Similar to displacement we have angular displacement and similar to velocity we have angular velocity. Similar to conservation of angular momentum, we have conservation of angular momentum in rotational motion. If only centripetal force is acting in a motion then it is called uniform circular motion. If there is another acceleration called tangential acceleration, we call the body is in non uniform circular motion and it is studied under rotational dynamics.

Problem

A tangential force of 10 newton is acting on a circular plate of radius 50 centimeter such that it can rotate about an axis perpendicular to the plane passing through the center. If moment of inertia is given to us, we need to measure the number of rotations it makes in the first six seconds. Problem is as shown in the diagram below.


Solution

We can measure the torque acting on a system as the product of moment of inertia and the angular acceleration. Thus we can get the value of angular acceleration. Using that data, we can measure the angular displacement using the corresponding equation of motion as shown in the diagram below. Angular velocity can be written as the component of angular frequency and the problem can be solved as shown in the diagram below.


Problem

A ceiling fan is rotating about about its kw  axis with a uniform angular velocity. Electric current is switched off and due to constant opposing torque its angular velocity to reduced to a new value and it happens in thirty rotations. We need to measure the number of rotations that the body takes before it comes to state of rest.


Solution

We can find angular acceleration using the initial and final angular velocity and angular displacement. We can also use third equation of motion and further simplify the given data as shown in the diagram below.


Problem

A simple pendulum is oscillating with an angular amplitude 90 degree. If the direction of resultant acceleration of the bob is horizontal at a point whee angle made by the string with vertical is. Problem is as shown in the diagram below.



Solution

Body in circular motion shall be having centripetal acceleration and we know the velocity at a point making some angle as given in the problem. We also know that the sin component of acceleration due to gravity and we can find the tangential acceleration using the respective formula. By dividing both the cases, we can get the answer as shown in the diagram below.


Problem

A circular disc of known radius and thickness has a known moment of inertia and it is perpendicular to the plane and passing though its center. It is melted and recasted into a solid sphere. We need to find the moment of inertia and the problem is as shown in the diagram below.


Solution

Even when the body is recasted its volume remains same and it can be equated as shown in the diagram and hence we can find the relation between two radius of the systems. By applying the rules and formula of moment of inertia, we can solve the problem as shown in the diagram below. Moment of inertia is a property of a body that opposes the rotational motion of the body.


Problem

Four uniform rods each of mass M and length L are connected in the form of a square. We need to find the moment of inertia of the system about the geometrical axis.


Solution

Moment of inertia of rod can be expressed in terms of its length and mass and we can find the moment of inertia of the rod about the given axis using parallel axes theorem as shown in the diagram below. Moment of inertia of the system is four times of each rod as all of them are identical. The solution is as shown in the diagram below.



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Rotational Dynamics Problems with Solutions Two

We are solving series of problems based on the concept rotational dynamics. Moment of inertia is a physical quantity that measures how difficult it is to rotate a body about a given axis. This is similar to mass of translatory motion. Moment of inertia can be defined as the summation of product of mass of each particle with  the square of the distance from the axis of rotation. It depends on the mass of the body and axis of rotation of the body. To over come this moment of inertia, we need to apply torque which means turning effect. It is cross product of distance of the body from axis of rotation and the force applied.

Problem

A mass 1.9 kilogram is suspended from a string of length half meter and it is at rest. Another body of mass 100 gram is moving horizontally strikes this mass and sticks to it. If the combined mass is just able to complete the circle, we need to find the initial velocity of the body. Problem is as shown in the diagram below.


Solution

The kind of collision after which both the bodies moves together is called inelastic collision and here both the bodies moves with common velocity. For this combination to move and complete vertical circular motion, it shall have some minimum velocity at the bottom and taking that into consideration, we can solve the problem as shown in the diagram below. We have applied law of conservation of linear momentum to the given data as shown.


Problem

Three point sized bodies each of same mass are fixed at the three corners of of triangle. We need to find the moment of inertia of the system about an axis passing through the center of frame and perpendicular to the frame. Problem is as shown in the diagram below.


Solution

Each particle is at a distance from the axis of rotation and it can be found using basic geometric principles. All of them are identical with axis and hence moment of inertia of the system is the sum of each of them. Problem is solved as shown in the diagram below.


Problem

Radius of gyration of a body is 18 centimeter when it is rotating about about an axis passing the center of mass of the body. If radius of gyration of the same body is 30 cm about a parallel axis to the first axis, then we need to find the perpendicular distance between two parallel axes.


Solution

Moment of inertia of a body depends on the axis of rotation and it changes with that. To find the moment of inertia of one axis that is parallel to another axis, we need to use parallel axes theorem. According to that theorem we can write statement as discussed in the chapter and further problem can be solved as shown in the diagram below.


Problem

Three identical rings each of mass m  and radius r are placed in the same plane such that each one ring is in touch with the other two. We need to find the moment of inertia of  the system about an axis passing through center of any one ring and the axis is perpendicular to the plane. Problem is as shown in the diagram below.


Solution

We know the moment of inertia of a ring about an axis passing through the center and perpendicular to the plane as a standard result. Other ring axis is a parallel axis to the first axis and their distance of separation is 2r. We can apply parallel axes theorem and find the moment of inertia of other two rings and the moment of inertia of the system is sum of all of them. Solution is as shown in the diagram below. 


Problem

The handle of a door is at a distance of 40 centimeter from the axis of rotation. If a force of 5 newton is applied on the handle at a known angle, we need to measure the torque generated and the problem is as shown in the diagram below.


Solution

We know that the torque is turning effect and it is the cross product of force applied to the distance of the particle from the axis of rotation. By applying that concept, we can solve the problem as shown in the diagram below.



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