Question

A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in figure. Then the acceleration of the body is:

A. $\dfrac{2mg}{2m+M}$

B. $\dfrac{2Mg}{2m+M}$

C. $\dfrac{2Mg}{2M+m}$

D. $\dfrac{2mg}{2M+m}$

Answer 1

Hint: Use free body diagram concept. Moment of inertia is given by $I=\dfrac{M{{R}^{2}}}{2}$. Tension T will be acted by string on mg. The direction of tension is in an upward direction so we take it negative. Whereas mg is acting downward so we take it as positive. Sum of forces must be equal to ma.  

Complete step by step solution:

We have a uniform disc having radius R and mass M which rotates about its axis. A string is wrapped over a disc ring as shown in the figure. To this string, a mass of m is tied to the free string. That is no other forces are applied to mass m other than tension by a string. Now we need to calculate the acceleration of the body.

Aim: Find the acceleration of the body.

As you can see in the given diagram, mass is handed to the string.

Now resolve forces on mass m.

We know that tension T is applied by a string on mass m. so the equation is,

$mg-T=ma................(a1)$

(Since mg is acting downward and T is acting upward. So the difference of it is equal to acceleration ma)

Now we know that a disc is rotating around its rim, so by the rotational motion concept.

Therefore the equation of rotational motion is given by,

$R.T=I\alpha ..............(1)$

Where, $R$ is the radius of the disc, $I$ is the moment of inertia, $T$ is the tension and $\alpha $ is angular acceleration.

We know that moment of inertia is given by,

$I=\dfrac{M{{R}^{2}}}{2}$

And angular acceleration is given by,

\[\alpha = \dfrac{a}{R}\]

So equation 1 can be written as,

$R.T=\dfrac{M{{R}^{2}}}{2}\times \dfrac{a}{R}$

Cancel R, we get,

$T=\dfrac{Ma}{2}$

Now put the above equation on equation (a1), we get

$mg-\dfrac{ma}{2}=ma$;

$mg=ma+\dfrac{Ma}{2}$

$mg=a\left( m+\dfrac{M}{2} \right)$

$a=\dfrac{2mg}{2M+m}$

Hence, the correct answer is option D.

Note: Tension is nothing but a kind of force. Tension words or tension concepts will definitely appear, where the string to thread kind of thing is used. Tension on mass m will always act upward. Angular acceleration is nothing but translation acceleration divided by distance.