Question

A solid cylinder, a circular disc, a solid sphere and hollow cylinder of the same radius are laced on an inclined plane. Which of the following will have maximum acceleration at the bottom of the plane?
A. Circular disc
B. Solid cylinder
C. Solid sphere
D. Hollow sphere

Answer 1

Hint: We need the expression for the acceleration of an object rolling down an inclined plane which is in terms of the moment of inertia of the object. By inserting the values of moment of inertia for various objects, we can find out which one has the maximum acceleration.
Formula used:
The acceleration of an object rolling down an inclined plane is given as
$a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{I}{m{r}^2}}}}$

Complete answer:
We are given a solid cylinder, a circular disc, a solid sphere and hollow cylinder which have the same radius and are allowed to roll down an inclined plane. We need to find out which of these objects has maximum acceleration when it reaches the ground.
The moment of inertia of various given objects is given as
A solid cylinder: $I={\displaystyle \frac{1}{2}}m{r}^2$
A circular disc: $I={\displaystyle \frac{1}{2}}m{r}^2$
A solid sphere: $I={\displaystyle \frac{2}{5}}m{r}^2$
A hollow cylinder: $I=m{r}^2$
Now we now that the acceleration of an object rolling down an inclined plane is given as
$a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{I}{m{r}^2}}}}$
Here I is the moment of inertia of the given object.
So, the corresponding acceleration of these objects can be written as
A solid cylinder: $a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{{\displaystyle \frac{1}{2}}m{r}^2}{m{r}^2}}}}={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{2}{3}}g\sin \theta$
A circular disc: $a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{{\displaystyle \frac{1}{2}}m{r}^2}{m{r}^2}}}}={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{1}{2}}}}={\displaystyle \frac{2}{3}}g\sin \theta$
A solid sphere: $a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{{\displaystyle \frac{2}{5}}m{r}^2}{m{r}^2}}}}={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{2}{5}}}}={\displaystyle \frac{5}{7}}g\sin \theta$
A hollow cylinder: $a={\displaystyle \frac{g\sin \theta }{1+{\displaystyle \frac{m{r}^2}{m{r}^2}}}}={\displaystyle \frac{g\sin \theta }{1+1}}={\displaystyle \frac{1}{2}}g\sin \theta$

So, the correct answer is “Option C”.

Note:
It should be noted that the solid sphere has the least value of moment of inertia compared to other given objects. This means that it is easier to roll the solid sphere as there is lesser resistance to rolling because of lesser moment of inertia. So, the acceleration of the solid sphere is found to be greater than all other given objects.