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# moment of inertia. how do?

A sphere with a diameter of 6.50 cm and a mass of 315 g is released from rest at the top of a 3.10 m, 17.0° incline. It rolls without slipping to the bottom. What is the sphere's angular velocity (in rad/s) at the bottom of the ramp? Assume that the local acceleration due to gravity is -9.80 m/s2 and that the moment of inertia of a sphere is 2/5 MR2.

### 2 Answers

- oldprofLv 71 decade agoFavorite Answer
The total energy before release is TE = mgh = mg S sin(theta) = GPE; where m = .315 kg, r = .065 m, S = 3.1 m, and theta = 17 deg. Then, through the conservation of energy law, the total energy at the bottom of the slope is.

TE = 1/2 mv^2 (1 + k), where k = 2/5 and v = V = tangential and linear speed as there is "no slippage."

Combine the two TE equations, 1.2 nv^2(1 + k) = mgS sin(theta), and solve for v = sqrt(2gS sin(theta)/(1 + k)) = ? mps ANS. g = 9.8 m/s^2, everything is given, you can do the math.

- IvanLv 61 decade ago
The energy balance eqn is

1/2 I w^2+1/2mv^2=mgh

but v=rw so

1/2Iw^2+1/2 m R^2.w^2=mgh

or 1/2 w^2(I+mR^2)=mgh

moment of Inertia of Sphere I=2/5mR^2

so

1/2 w^2(2/5mR^2+mR^2)=mgh

w^2=2gh/(2/5+1)R^2=10gh/7R^2=4.11

w=2.03rad/s

IVAN