Question

A uniform-density wheel of mass 10 kg and radius 0.36 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of 14 N for 0.76 s. (a) What is the final angular speed? radians/s (b) What was the average angular speed? radians/s (c) Through how big an angle did the wheel turn? radians/s (d) How much string came off the wheel?

Answer 1

Answer:

(a) 5.91 rad/s

(b) 2.96 rad/s

(c) 2.25 rad

(d) 0.81 m

Explanation:

The torque generated by tension force from the string is:

T = FR = 14*0.36 = 5.04  Nm

This torque would then create an angular acceleration on the uniform-density wheel with moments of inertia of

$I = 0.5mR^2 = 0.5*10*0.36^2 = 0.648kgm^2$

$\alpha = \frac{T}{I} = \frac{5.04}{0.648}=7.78rad/s^2$

(a) The wheel turns for 0.76s, this means the final angular speed is

$\omega_f = t\alpha = 0.76*7.78 = 5.91 rad/s$

(b) Since the force is constant, the torque is also constant and so is the angular acceleration. This means angular speed is rising at a constant rate. That means the average angular speed is half of the final speed

$\omega_a = 0.5\omega_f = 0.5*5.91 = 2.96 rad/s$

(c) The total angle that the wheel turns is the average angular speed times time

$\theta = t\omega_a = 2.96*0.76 = 2.25 rad$

(d) The string length coming off would equal to the distance swept by the wheel

$d = R\theta = 0.36*2.25 = 0.81 m$

Answer 2

The final angular speed of this uniform-density wheel of mass is equal to 5.91 radians/s.

Given the following data:

Mass = 10 kg.

Radius = 0.36 m.

Initial velocity = 0 m/s (since it's starting from rest).

Force = 14 Newton.

Time = 0.76 seconds.

How to calculate the final angular speed.

First of all, we would determine the torque produced due to the tensional force that is acting on the string by using this formula:

$\tau = Fr\\\\\tau = 14 \times 0.36$

Torque = 5.04 Nm.

Also, we would determine the moment of inertia by using this formula;

$I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 10 \times 0.36^2\\\\I=5 \times 0.1296$

I = 0.648 $kgm^2$

Next, we would determine the angular acceleration by using this formula;

$\tau=\alpha I\\\\\alpha =\frac{\tau}{I} \\\\\alpha =\frac{5.04}{0.648}\\\\\alpha = 7.78 \;rad/s^2$

Now, we can calculate the final angular speed:

$\omega_f = t\alpha \\\\\omega_f = 0.76 \times 7.78\\\\\omega_f = 5.91 \;rad/s$

How to calculate the average angular speed.

$\omega_A = \frac{1}{2} \omega_f\\\\\omega_A = \frac{1}{2} \times 5.91\\\\\omega_A =2.96\;rad/s$

How to calculate the angle.

$\theta = t\omega_A \\\\\theta = 0.76 \times 2.96$

Angle = 2.25 rad.

In order to calculate the length of the string that came off the wheel, we would determine the distance swept by the wheel:

$d=r\theta\\\\d=0.36 \times 2.25$

d = 0.81 meter.

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