## Wednesday, December 14, 2016

### 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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