Question

A hollow sphere of mass M and radius R rolls on a horizontal surface without slipping such that the velocity of its centre of mass is V.The total kinetic energy of the sphere is ​

Answer 1

If a solid sphere of mass M and radius R is rolling perfectly on a rough horizontal surface, what is the percentage of the rotational kinetic energy?

Te total kinetic energy (KE) of an object is given by:

KE=12mv2+12Iω2

where I is the moment of inertia of the object and ω its angular momentum. For a solid sphere, the moment of inertia is given by:

I=25mr2

angular velocity would be derived from its radius and linear velocity:

ω=vr

so the total equation would be:

KE=12mv2+1225mr2vr2

=12mv2+210mv2

To get the percentage attributed to rotational energy, we’d divide the rotational part of the energy by the total energy:

210mv212mv2+210mv2=210mv2710mv2=27

Answer 2

Answer:

The total kinetic energy of the sphere is 5/6 mv^2.

Explanation:

As per the data given in the questions

We have to calculate total kinetic energy of the sphere.

As per the question

It is given that

A hollow sphere of mass M and radius R rolls on a horizontal surface without slipping such that the velocity of its center of mass is V.

According to angular momentum theory

As we know that Translational kinetic energy of sphere =$K_{T}$ = $\frac{1}{2}$×$mv^{2}$

Where I is the moment of inertia of object and w is the angular momentum.

For hollow sphere I =1/3 mv^2

Rotational kinetic energy of sphere=$K_{R}$ = $\frac{1}{2}$×I$w^{2}$=$\frac{2}{2}*\frac{mv^{2} }{3} *w^{2}$

For rotating v=r w

So,$K_{R}=\frac{mv^{2} }{3}$

Total Kinetic energy = Translational kinetic energy+ Rotational kinetic energy

⇒Total Kinetic energy = $K_{T}+K_{R}$

⇒$K.E_{Total}=(\frac{1}{2}+\frac{1}{3}) mv^{2}$

⇒$K.E_{Total}$ = $\frac{5}{6}mv^{2}$

Hence, the total kinetic energy of the sphere is 5/6 mv^2.

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