Question

A wheel of mass 50 kg has a radius of 0.4 m. It is making 480 rpm. What is the

torque necessary to bring it to rest in 40 seconds?
Solution:

Answer 1

-10.048 N m

Step-by-step explanation:

Answer 2

The torque necessary to bring the wheel to rest in 40 seconds is 10.4 N·m

Step-by-step explanation:

The question is with regards to rotational motion

The rotary motion parameters are;

The mass of the wheel = 50 kg

The radius of the wheel = 0.4 m

The rate of rotation of the wheel = 480 rpm

The time in which the wheel is to be brought to rest = 40 s

The rotational rate of the wheel in rotation per second is given as follows;

480 r.p.m = 480 r.p.m × 1 minute/(60 seconds) = 8 revolution/second

1 revolution = 2·π radians

Therefore, we have the angular velocity, ω, given as follows;

ω = 2·π × 8 revolutions/second ≈ 50.3 rad/s

The angular acceleration, α, is given as follows;

`\alpha = (\Delta \omega)/(\Delta t) = (\omega _2 - \omega_1)/(t_2 - t_1)`

Whereby the wheel is brought to rest from its initially constant rotational motion in 40 seconds, we have;

ω₁ ≈ 50.3 rad/s, ω₂ = 0 rad/s, and t₂ - t₁ = 40 seconds

Plugging in the values for the variables of the equation for the angular acceleration, "α", we get;

`\alpha = (0 - 50.3 \ rad/s)/(40 \ s) \approx 1.3 \ rad/s^2`

The torque on the wheel, τ, is given as follows;

τ = m·r²·α

Where;

m = The mass of the object = 50 kg

r = The radius of the wheel = 0.4 m

α = The acceleration of the wheel ≈ 1.3 rad/s²

Therefore;

τ = 50 kg × (0.4 m)² × 1.3 rad/s² ≈ 10.4 N·m

The torque necessary to bring the wheel to rest in 40 seconds = τ ≈ 10.4 N·m.