__Question 7.18__:

__Solution__:**Answer:**

**(a)**

**(a)**Yes

Mass of the sphere =

*m*

Height of the plane =

*h*

Velocity of the sphere at the bottom of the plane =

*v*

At the top of the plane, the total energy of the sphere = Potential energy =

*m*g

*h*

At the bottom of the plane, the sphere has both translational and rotational kinetic energies.

Hence, total energy = (1/2)mv

^{2}+ (1/2) I ω

^{2}

Using the law of conservation of energy, we can write:

(1/2)mv

^{2}+ (1/2) I ω

^{2}= mgh

For a solid sphere, the moment of inertia about its centre, I = (2/5)mr

^{2}

Hence, equation (

*i*) becomes:

(1/2)mr

^{2}+ (1/2) [(2/5)mr

^{2}]ω

^{2}= mgh

(1/2)v

^{2}+ (1/5)r

^{2}ω

^{2}= gh

But we have the relation, v = rω

∴ (1/2)v

^{2}+ (1/5)v

^{2}= gh

v

^{2}(7/10) = gh

v = √

*(10/7)gh*

Hence, the velocity of the sphere at the bottom depends only on height (

*h*) and acceleration due to gravity (g). Both these values are constants. Therefore, the velocity at the bottom remains the same from whichever inclined plane the sphere is rolled.

**(b)**Yes

**(c)**On the smaller inclination

Consider two inclined planes with inclinations

*θ*

_{1 }and

*θ*

_{2}, related as:

θ

_{1}<

*θ*

_{2}

The acceleration produced in the sphere when it rolls down the plane inclined at

*θ*

_{1}is:

g sin

*θ*

_{1}

The various forces acting on the sphere are shown in the following figure.

*R*

_{1}is the normal reaction to the sphere.

Similarly, the acceleration produced in the sphere when it rolls down the plane inclined at

*θ*

_{2}is:

g sin

*θ*

_{2}

The various forces acting on the sphere are shown in the following figure.

*R*

_{2}is the normal reaction to the sphere.

θ

_{2}>

*θ*

_{1}; sin

*θ*

_{2}> sin

*θ*

_{1}... (

*i*)

∴

*a*

_{2}>

*a*

_{1 … }(

*ii*)

Initial velocity,

*u*= 0

Final velocity,

*v*= Constant

Using the first equation of motion, we can obtain the time of roll as:

*v*=

*u*+

*at*

∴ t ∝ (1/α)

For inclination θ

_{1}: t

_{1}∝ (1/α

_{1})

For inclination θ

_{2}: t

_{2}∝ (1/α

_{2})

From above equations, we get:

*t*

_{2 }<

*t*

_{1}

Hence, the sphere will take a longer time to reach the bottom of the inclined plane having the smaller inclination.

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