Question

A solid sphere with mass M=4.2kg and radius R=0.25m rolls across
the floor without slipping. If the sphere has a totoal kinetic
energy of K=6.5J **what is the angular speed?**

Answer #1

As we know that the linear kinetic energy is defined by K_linear
= (1/2)mv^2, where

m = mass of the body in motion

v = velocity of the body in motion

Rotational kinetic energy is defined by K_rotational = 1/2Iω^2,
where

I = mass moment of inertia of the body in motion

ω = rotational velocity of the body in motion

For solid sphere, I=2/5mr^2

where r = the radius of the sphere

Put this in the above –

We get, K_rotational = 1/2(2/5mR^2)ω^2 = 1/5m(R^2)(ω^2) = (1/5)mv^2 [since v = Rω]

So, total kinetic energy, KE = (1/2)mv^2 + (1/5)mv^2 = 0.7*
mv^2

Now given that total kinetic energy = 6.5 J

So, 0.7* mv^2 = 6.5

=> 0.7*4.2*v^2 = 6.5

=> v = 1.49 m/s

Now, v = R* ω

=> ω = v/R = 1.49/0.25 = 5.96 rad/s.

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