Question

A constant torque of 26.6 N · m is applied to a grindstone for which the moment of inertia is 0.162 kg · m2 . Find the angular speed after the grindstone has made 11.8 rev. Assume the grindstone starts from rest.

Answer 1

Answer:

156.1 rad/s = 24.8 rev/s

Explanation:

Torque = Momentum of inertial × radial acceleration = Iα

τ = 26.6 N.m

I = 0.162 kg.m²

26.6 = 0.162 × α

α = 164.2 rad/s²

Using equations of motion,

θ = 11.8 rev = 11.8 × 2π = 74.2 rad

w₀ = 0 rad/s (since the grindstone starts from rest)

w = ?

α = 164.2 rad/s²

w² = w₀² + 2(α)(θ)

w² = 0² + (2×164.2)(74.2)

w = 156.1 rad/s = 24.8 rev/s

Hope this Helps!!!

Answer 2

Answer:

Explanation:

wi = 0 rad/s (starts from rest)

Torque, T= 26.6 N.m

Moment of inertia, I = 0.162 kg.m2

Theta = 11.8 rev

Converting rev to rad,

1 rev = 2pi rad

11.8 rev × 2pi rad/1 rev

= 74.14 rad

Using the formula,

T = I × a

Where,

a = angular acceleration

a = 26.6/0.162

= 164.2 rad/s^2

Using equations of circular rmotion,

wf^2 = wi^2 + 2a × theta

= 0 + 2 × 164.2 × 74.14

= 12174.05

wf = sqrt(12174.05)

= 110.34 rad/s.