Question

Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 60 N and that the pressured water provides a frictionless support. How long will it take her to rotate the sphere one time, starting from rest?

Answer 1

25.06s

Step-by-step explanation:

Remaining part of the question.

(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)

Solution:

F = 60N

r = 90cm = 0.9m

M = 8200kg

Moment of inertia for a sphere (I) = ⅖mr²

I = ⅖ * m * r²

I = ⅖ * 8200 * (0.9)²

I = 0.4 * 8200 * 0.81

I = 2656.8 kgm²

Torque (T) = Iα

but T = Fr

Equating both equations,

Iα = Fr

α = Fr / I

α = (60 * 0.9) / 2656.8

α = 0.020rad/s²

The time it will take her to rotate the sphere,

Θ = w₀t + ½αt²

Angular displacement for one revolution is 2Π rads..

θ = 2π rads

2π = 0 + ½ * 0.02 * t²

(w₀ is equal to zero since sphere is at rest)

2π = ½ * 0.02 * t²

6.284 = 0.01 t²

t² =6.284 / 0.01

t² = 628.4

t = √(628.4)

t = 25.06s