Gravitation Problems with Solutions Four

Gravitational field intensity is the force experienced by a unit massed particle when it is placed in the gravitational field of another particle of known mass and its direction is similar to gravitational force. Planetary motion is identified basing on Kepler laws. According to first law, planets take elliptical path around a star. According to second law, areal vector of a planet sweeps equal area in in equal intervals of time. Basing on this it can be said that when a planet comes near to the star its velocity will be more and vice versa.

Problem

Infinite number of masses each of mass one kilogram are placed along a straight line at a uniform distance of separation from the origin. We need to find the gravitational intensity at the origin due to all the other masses and the problem is as shown in the diagram below.

Solution

We know that gravitational force acts on the particle at the origin and as we need to find the intensity, we shall consider a unit mass at the origin. Each particle on the x axis applies gravitational force of attraction on it  and the total force is the vector sum of all the forces acting on it. Solution is as shown in the diagram below. Sum of the terms can be found using the geometrical progression sum using the standard formula as shown in the diagram below.

Problem

Gravitational field due to a mass at a distance due to a given mass as shown in the equation below. Assuming gravitational potential to be zero at infinity, we need to find its value at a given point and the problem is as shown in the diagram below.

Solution

We can express the gravitational potential in terms of the intensity as shown in the diagram below. Potential is the work need to be done to bring a mass of unit mass from infinite to a particular location. Problem can be further simplified as shown in the diagram below.

Problem

Two spheres are separated by a known distance and we need to know the angle which the strings will make with the vertical  due mutual attraction of the spheres and the problem as shown in the diagram below.

Solution

As the strings turn tight, tension is developed in the string and it can be resolved into components. By dividing the two equations, we can solve the problem as shown in the diagram below.

Problem

The value of acceleration due to gravity on the surface of the earth is given. If the point where we need to find the acceleration due to gravity from the surface of the earth and we need to find that height and the problem is as shown in the diagram below.

Solution

We can just apply the acceleration due to gravity variation with respect to height and simplify the equation as shown in the diagram below.

Problem

Three particles are placed at the three corners of a equilateral triangle of side a. We need to find the work done in increasing the side length to double to its initial value. Problem is as shown in the diagram below.

Solution

Each particle will get some potential energy with respect to other and as the separation increases potential energy also increases. Work done is the difference between initial and final potential energies of the system. Solution is as shown in the diagram below.

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Gravitation Problems with Solutions Three

Gravitational force is one of the fundamental and weakest force of the nature and it exists between every two bodies as every body is having some masses. Gravitational force is due to the mass of the bodies and it is independent of the medium. It is a long distance ranged force and it acts between the particles separated by a very large distance. When compared with electromagnetic and nuclear force, it is just minimum and it gives the stability to the system. The  earth is revolving around the sun due to this gravitational force and it acts like the centripetal force. We are solving series of problems on this topic and this post is the third one in the series.

Problem

We would like to find the acceleration due to gravity of the earth at a height equal to half the radius of the earth above the earth. Problem is as shown in the diagram below.

Solution

While solving this case, approximation can not be applied as the height is significant when compared with the radius of the earth. By applying the newton's law of gravitation, we can find the gravity at a given height as shown in the diagram below.

Problem

Two equal masses are hung from a balance whose scale pans differ in height by h. If density of the earth is given to us, we need to find the error in the weighting. Problem is as shown in the diagram below.

Solution

It is shown in the diagram that from the reference point, both the pans are at different heights and hence they have a difference in the heights as given in the problem. Error in the reading is the difference in its weights. Variation of heights is small and hence approximation can be applied and the problem can be solved as shown in the diagram below.

Problem

A small body of super dense material whose mass is twice the mass of the earth but size is very small when compared with the earth. If it starts from the state of rest, we need to know its location after a specified time. The problem is as shown in the diagram below.

Solution

For every action there is equal and opposite reaction. As the body comes downward, the earth moves upward. The two meet at the center of mass of the system. We can further solve the problem as shown in the diagram below.

Problem

We need to find the height from the surface of the earth where weight of the body is similar to the weight at a depth from the surface of the  earth. Problem is as shown in the diagram below.

Solution

We need to apply the formula for the acceleration due to gravity at a height and at a depth and we need to just equate both of them. The problem is as shown in the diagram below.

Problem

We need to find how fast the earth has to rotate so that apparent weight of a body becomes zero at the equator. Problem is as shown in the diagram below.

Solution

We know how acceleration due to gravity varies with latitude. We know that at equator it is minimum and as we are moving to wards poles, it becomes maximum. By writing the formula at the equator and further equating weight at that point zero, we can find the angular velocity as seventeen times the normal velocity and the detailed solution is as shown in the diagram below.

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Gravitation Problems with Solutions Two

We are  solving series of problems about a physics concept called gravitation. We are predominately are solving problems based on force of attraction between two bodies. It is always attractive nature and it is never repulsive. This force exists between every two bodies with out any escape as they are all having some masses. It is independent of medium and the proportionality constant is called universal gravitational constant. The variation of acceleration due to gravity varies with height and depth from the surface of the earth. The loss of acceleration due to gravity is quicker with height when compared with the depth.

Problem

Here in this problem we need to know the variation of acceleration due to gravity with the depth from the surface of the earth. Problem is as shown in the diagram below.

Solution

Let us consider the location where we need to measure the acceleration due to gravity is at a depth d from the surface of the earth. From the center of the earth then it is at a distance of R-d. So on the given mass, entire mass of the earth is not applying the gravitational force of attraction, rather a portion of the mass under it.

So we shall count only that portion of the mass. So we express it in terms of volume and density and further simplify as shown in the diagram below.

Problem

Three uniform spheres each of identical mass and radius are kept in such a way that each one touch the other two spheres. We need to find the total gravitational force acting on each sphere and the problem is as shown in the diagram below.

Solution

On any one sphere, there is gravitational force due to the other two and to find the resultant force, we need to use vector laws of addition. Distance between any two spheres centers is nothing but twice the radius of any sphere.

We can find the resultant as shown in the diagram below and its direction also can be identified. Complete and detailed solution is given below.

Problem

Four identical masses are placed at the four corners of a square and we need to find the effective gravitational force at any one corner and the problem is as shown in the diagram below.

Solution

On any one particle, there is gravitational force of attraction due to the other two. We can find each force using the newton's law of gravitation. The diagram is as shown below.

It can be further solved as shown below. Two of the total three forces are equal in magnitude and perpendicular to each other. Their resultant is along the direction of the third force and hence we can find the total resultant as shown in the diagram below.

Problem

A satellite is going along an elliptical path around the earth. We need to find the relation between rate of area swept by it in terms of its radius and the problem is as shown in the diagram below.

Solution

According to second law of Kepler, the velocity vector of a planet sweeps equal area in equal intervals of time. Taking that into consideration, we can solve the problem as shown in the diagram below.

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Gravitation Problems with Solutions One

Gravitational force exists between every two particles having some mass and it is directly proportional to the product of their masses and inversely proportional to the square of distance of separation. It is independent of medium between them. If number of bodies are present around any body, the total gravitational force is the vector sum of all the existing forces. We also need to know that the presence of third body is not going to effect the gravitational force between the other two bodies. Acceleration due to gravity varies basing on the position of the body from the surface of the earth and depth of the body from the surface of the earth. It also depends on the location of the body with respect to its location of equator.

Problem

It is given in the problem that the gravitational force between two objects at a separation of one meter is given to us and we need to find the mass of each object.

Solution

We can apply law of gravitational force to know the force between the two bodies and the force is given in the problem and their separation is also given in the problem. By applying them in the formula and by applying the constant value of universal gravitational constant, we can solve the problem as shown in  the diagram below.

Problem

It is given in the problem that mass of one ball is four times the mass of the other body and their separation is given as 10 centimeter. The gravitational force between them is given to us and we need to find the mass of each particle.

Solution

We need to apply once again newton's law of gravitation as shown in the diagram below. Universal gravitational constant is a proportional constant and it is constant over the entire universe. Data can be further simplified and problem can be solved as shown in the diagram below.

Problem

It is given in the problem that three spherical balls of mass 1,2 and 3 kilogram are placed at the corners of an equilateral triangle of side one meter. We need to find the effective gravitational force acting on the body of mass one kilogram.

Solution

Let us consider three particles at the three corners of the equilateral triangle as shown in the diagram below. On one kilogram  particle there is now gravitational attractive force acting due to other two bodies of masses 2 and 3 kilogram. Their directions are along the line joining them and their resultant can be found using the parallelogram law of vectors.

It can be further simplified as shown in the diagram below. Vector law of addition is applied and we know the angular separation between the bodies and it can be further simplified as shown below.

Problem

We need to know that how a particle of known mass has to be divided into two parts such that the gravitational force between them is maximum and the problem is as shown in the diagram below.

Solution

We can find the gravitational force between any two bodies using the Newton's law of gravitation. We also need to know that if any function has to be maximum, its corresponding differentiation has to become zero as shown in the diagram below. By applying the rules of differentiation and simplifying it further, we can solve the problem as shown in the diagram below.

Problem

It is given that two particles of known mass are separated by a certain distance. We need to know at what distance from the body of mass one kilogram, the gravitational force becomes zero and the problem is as shown in the diagram below.

Solution

We shall imagine any particle of certain mass between the two bodies along the line joining them. On the third body, both the bodies applies force of attraction towards them and hence both of them on the third body are in opposite direction. At any instant, the resultant force becomes zero and the point becomes a neutral point. The problem is solved as shown in the diagram below.

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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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