Rotation problem (I am completely lost please help!)

Question

A girl pushes on a large stone sphere. The sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a on the sphere tangent to its surface with a steady force of F= 25 N and that the pressured water provides a frictionless support.

1. How long will it take her to rotate the sphere one time, starting from rest?

Answer 1

Given r = 90 cm = 0.9 m, F = 25 N, m = 8200 kg, θ = 2π rad (one full rotation)

In rotational motion equations

Linear variables get replaced by angular variables

x (m) → θ (rad), v (m/s) → ω (rad/s), a (m/s²) → α (rad/s²)

F (N) → τ (N⋅m) where τ = torque (force × lever arm)

And the familiar equation F = ma

Becomes τ = I α where I = moment of inertia

To solve for t, we use the rotational kinematics equation

θ = ω₀t + ½αt²

where ω = angular velocity and α = angular acceleration

(Note that this is analogous to the linear kinematics equation (x = v₀t + ½at²)

since we’re starting from rest, ω₀ = 0 and

θ = ½αt²

So t = √(2θ/α) and since θ = 2π rad

t = √(4π/α)

To find t, we need to calculate α

τ = Iα where

τ = rF and I = ⅖mr² (the moment of inertia of a solid sphere)

so

α = τ/I = rF/(⅖mr²) = 2.5F/(mr) = 2.5(25 kg m/s²)/[(8200 kg)(0.9 m)] rad/s²

α ≈ 8.47 × 10⁻³ rad/s²

So

t = √(4π/α) = √[(4π rad)/(8.47 × 10⁻³ rad/s²)] ≈ 38.5 s

Answer

It will take her about 38.5 seconds to rotate the sphere one time.