A rubber ball mass = 50 g radius r = 0.25 m, is spun by a force F = 100 N tangen

Question

A rubber ball mass = 50 g radius r = 0.25 m, is spun by a force F = 100 N tangent to its surface. The ball is spun at sea level (P_atm = 1.01 times 10^3 Pa). You may treat the ball as a hollow sphere, I_ms =. A) Find the magnitude of the angular acceleration of the ball. B) Suppose the ball were placed deep under water such that the pressure, P, was 9 times greater than the pressure in air. How much would the volume of our ball change? Since our sphere is hollow, its effective bulk modulus is small- B = 3.5 times 10^6 Pa C) The same force from part A is now applied. What is the new angular acceleration? You may ignore drag effects from the water.

Explanation / Answer

a) alpha is the angular accelaration

Torque is T = r*F = I*alpha

r = 0.25 m

F = 100 N

I = (2/3)*m*r^2 = (2/3)*50*10^-3*0.25^2 = 0.00208 kg-m^2 = 2.08*10^-3 kg-m^2

then

alpha = (r*F)/I = (0.25*100)/(0.00208) = 12019 rad/s^2

B) bulk modulus is B = dp /(dV/V)

dP = 9P -P = 8P = 8*1.013*10^5 Pa = 8.104*10^5 Pa

dV/V = dP/B = 8.104*10^5/(3.5*10^6) = 0.2315

change in volume is dV = 0.2315*V = 0.2315*(4/3)*pi*r^3 = 0.2315*(4/3)*3.142*0.25^3 = 0.01515 m^3

V2-V1 = 0.01515

V2= 0.01515 + ((4/3)*3.142*(0.25^3))

V2 = 0.0806 m^3

(4/3)*3.142*r^3 = 0.0806

r = 0.27 m

Now using Torque is T = r*F = I*alpha

0.27*100 = (2/3)*50*10^-3*0.27^2*alpha

alpha = 11111.12 rad/s^2

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