(13.4.40) A solid sphere of diameter 30.0 cm is pivoted at a point on its surfac

Question

(13.4.40) A solid sphere of diameter 30.0 cm is pivoted at a point on its surface. It is held so that the diameter passing through both the pivot and the center of the sphere makes a 0.15 radian angle with the vertical. The sphere is released at time t = 0.10 s. (a) What is the period of the resulting oscillation? (b) What is the angle theta between the radius to the pivot and the vertical as a function of time? (c) What is the angular velocity of the sphere as a function of time? (Remember that angular velocity is d theta/dt, which is a totally different quantity than the angular frequency omega.) (d) What is the speed of the point at the bottom of the sphere at time t = 0.40 s?

Explanation / Answer

Summing torques,

-m g (d/2) theta = I alpha

As by parallel axis theorem,

I = 2/5 m (d/2)^2 + m (d/2)^2

I = 7/20 m d^2

Thus,

-m g (d/2) theta = (7/20 m d^2) alpha

Solving for alpha,

alpha = -(10/7) (g/d) theta

Thus,

T = 2 pi sqrt( 7d / 10g)

Thus,

T = 0.9198 s   [ANSWER]

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Also,

w = sqrt[(10/7)(g/d)] = 6.8313 rad/s

Thus,

theta = 0.15cos(6.8313 rad/s * t)   [ANSWER]

***********************

As angular velocity = dtheta/dt,

angular velocity = -1.0247sin(6.8313 rad/s * t)   [ANSWER]

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Here,

v = d * angular velocity

at t = 0.40 s,

v = 0.122 m/s   [ANSWER]

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