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A 6.5kg ball with radius 8 cm rolls without Stopping down a Lone at 5m/s. calculate its kenetic energy (I= ² / 3 MR²)​

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The kinetic energy of a rolling ball can be calculated as the sum of its translational and rotational kinetic energies:

KE = 1/2 mv^2 + 1/2 Iω^2

where KE is the total kinetic energy, m is the mass of the ball, v is its velocity, I is the moment of inertia of the ball, and ω is its angular velocity.

To calculate the moment of inertia I of the ball, we can use the formula you provided:

I = (2/3)MR^2

where M is the mass of the ball and R is its radius.

Substituting the given values, we have:

M = 6.5 kg
R = 0.08 m

So, the moment of inertia of the ball is:

I = (2/3)(6.5 kg)(0.08 m)^2 ≈ 0.027 kg m^2

Now we can calculate the translational kinetic energy of the ball:

KE_trans = 1/2 mv^2 = 1/2 (6.5 kg)(5 m/s)^2 = 81.25 J

Next, we need to calculate the angular velocity ω of the ball. For a rolling ball without slipping, the tangential velocity v and the angular velocity ω are related by:

v = Rω

So, we have:

ω = v/R = 5 m/s / 0.08 m = 62.5 rad/s

Finally, we can calculate the rotational kinetic energy of the ball:

KE_rot = 1/2 Iω^2 = 1/2 (0.027 kg m^2)(62.5 rad/s)^2 ≈ 52 J

Therefore, the total kinetic energy of the rolling ball is:

KE = KE_trans + KE_rot = 81.25 J + 52 J ≈ 133.25 J

So, the kinetic energy of the rolling ball is approximately 133.25 J.
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