## Friday, December 16, 2016

### 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.

Related Posts