NSEP 2015 Problem 2 | Rotational Motion

A body of mass m and radius R rolling horizontally without slipping at a speed v climbs a ramp to a height $\frac{3v^2}{4g}$. The rolling body can be
(a) A Sphere
(b) A circular Ring
(c) A Spherical Shell
(d) A Circular Disc

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) A Circular Disc

Notice that It's a motion without slipping. This implies that the point of contact between the body and the ground has zero relative velocity.

If the relative velocity between the ground and object is 0, then we can use the relation $v = \omega R$, where $\omega $ is the angular velocity.

We know the kinetic energy for the rotational motion is,

$$K_{rot} = \frac{1}{2}I^2 \omega = \frac{Iv^2}{2R^2}$$

Hence, the total kinetic energy is,

$K_{\text {tot }}=K_{\text {linear }}+K_{\text {rot }}$=$\frac{1}{2} m v^{2}+\frac{I v^{2}}{2 R^{2}}$=$\frac{m v^{2}}{2}\left(1+\frac{I}{m R^{2}}\right)$

The object can only go upto the height $\frac{3v^2}{4g}$. At that height it's kinetic energy will be zero and all the energy will be potential energy. So, the potential energy at that height is,

$$ V = mgh = mg\frac{3v^2}{4g} = \frac{3}{4}mv^2 $$

Using the conservation of energy,

$\frac{m v^{2}}{2}\left(1+\frac{I}{m R^{2}}\right)=\frac{3}{4} m v^{2} \rightarrow I$

$=\frac{1}{2} m R^{2}$

We know this is the Moment of Inertia of a Circular disc. Hence, the object is a circular disc.

There is a nice way to remember those moment of inertia values of some standard bodies. , Maybe we will see that in some later blog.

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