A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane: 1. The hollow sphere reaches the bottom first. 2. The solid sphere reaches the bottom with greater speed. 3. The solid sphere reaches the bottom with greater kinetic energy. 4. The two spheres will reach the bottom with same linear momentum.

2. The solid sphere reaches the bottom with greater speed. Explanation: Acceleration of a sphere on the incline plane is given by: a= g 1+ I_ COM mr^2 ICOM for a solid sphere =2 5 mr^2 So, a= g 1+ 2mr^2 5mr^2 =5 7 g ICOM for a hollow sphere =2 3 mr^2 So, a'= g 1+ 2mr^2 3mr^2 =3 5 g The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

A hollow sphere and a solid sphere having same mass and same radi…

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Question

A hollow sphere and a solid sphere having same mass and same radii are rolled down a rough inclined plane:

The hollow sphere reaches the bottom first.
The solid sphere reaches the bottom with greater speed.
The solid sphere reaches the bottom with greater kinetic energy.
The two spheres will reach the bottom with same linear momentum.

✓

Answer

The solid sphere reaches the bottom with greater speed.

Explanation:

Acceleration of a sphere on the incline plane is given by:

$\text{a}=\frac{\text{g}\sin\theta}{1+\frac{\text{I}_{\text{COM}}}{\text{mr}^2}}$

I_COM for a solid sphere $=\frac{2}{5}\text{mr}^2$

So, $\text{a}=\frac{\text{g}\sin\theta}{1+\frac{2\text{mr}^2}{5\text{mr}^2}}=\frac{5}{7}\text{g}\sin\theta$

I_COM for a hollow sphere $=\frac{2}{3}\text{mr}^2$

So, $\text{a}'=\frac{\text{g}\sin\theta}{1+\frac{2\text{mr}^2}{3\text{mr}^2}}=\frac{3}{5}\text{g}\sin\theta$

The acceleration of the solid sphere is greater; therefore, it will reach the bottom with greater speed.

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