# A hollow spherical shell of mass 2.00 kg and radius 20 cm rolls without slipping down a...

## Question:

A hollow spherical shell of mass 2.00 kg and radius 20 cm rolls without slipping down a {eq}35.0^\circ {/eq} slope. Find the speed of the sphere after it travels 8 m down the slope, by using energy conservation. The moment of inertia of a hollow sphere about its center of mass is {eq}I = (2/3) MR^2 {/eq}.

## Speed of a Spherical Shell Rolling Down an Incline

When an object of mass *m* and moment of inertia *I* dolls down an inclined plane it gets converted some of its gravitational potential energy as kinetic energy of the rolling object. The decrease in potential energy of the rolling object will be proportional to the decrease in its vertical level from its initial position. The rolling object has linear kinetic energy and rotational kinetic energy. According to the law of conservation of energy, the decrease in potential energy must be equal to the gain in total kinetic energy of the rolling object.

## Answer and Explanation:

**Given points**

- Mass of the spherical shell
*M*= 2.00 kg - Radius of the spherical shell
*R*= 0.20 m - Angle of inclination of the plane {eq}\theta = 35^0 {/eq}

- Distance rolled down on the inclined plane
*L*= 8 m - Moment of inertia of the spherical shell {eq}I = \dfrac { 2 } { 3 } M R^2 {/eq}

When the sphere rolled down a distance of *L* meters , the vertical level of the spherical shell decreases from its initial position.

The decrease in vertical level of the spherical shell {eq}h = L \sin ( 35 ) \\ = 8.00 \times \sin( 35) \\ = 4.589 \ \ m {/eq}

So the decrease in potential energy of the spherical shell {eq}\Delta U = M g h {/eq}

This must be equal to the gain in kinetic energy of the spherical shell.

Let {eq}v, \ \ \omega {/eq} be the velocity of the center of mass of the spherical shell and the angular velocity of the spherical shell after it has rolled down *8* meters

Also we have the relation {eq}v = R \omega {/eq}

The energy conservation can be written as {eq}M g h = \dfrac { 1 } { 2 } \times \dfrac { 2 } { 3 } M R^2 \times ( \dfrac { v } { R } )^2 + \dfrac { 1 } { 2 } M v^2 \\ \implies gh = \dfrac { v^2 } { 3 } + \dfrac { v^2 } { 2 } {/eq}

Sp speed of the sphere at the given point {eq}v = \sqrt { \dfrac { gh } { \dfrac { 1 } { 3 } + \dfrac { 1 } { 2 } } } \\ = \sqrt { \dfrac { 9.8 \times 4.589 } { \dfrac { 1 } { 3 } + \dfrac { 1 } { 2 } } } \\ = 7.346 \ \ m/s {/eq}

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