Question

A wheel accelerates uniformly from rest to an angular speed of 25 rad/s in 10 s.(a) Find the angular acceleration of the wheel. (b) Find the tangential and radial acceleration of a point 10cm from the wheel’s center. (c) How many revolutions has the wheel turned during this time interval? (d) Then, find the wheel’s angular deceleration if it comes to a full stop after 5 rev

Answer 1

(a) Angular acceleration

The initial angular speed of the wheel is $\omega_i=0$, the final angular speed is $\omega_f=25 rad/s$, and the time taken is t=10 s. Therefore, the angular acceleration of the wheel is given by:

$\alpha=\frac{\omega_f-\omega_i}{t}=\frac{25 rad/s-0}{10 s}=2.5 rad/s^2$

(b1) Tangential acceleration

The tangential acceleration is given by the product between the angular acceleration $\alpha$ and the distance from the wheel's center r. In this case, $\alpha=2.5 rad/s^2$  and $r=10 cm=0.1 m$, therefore the tangential acceleration is

$a_t=\alpha r=(2.5 rad/s^2)(0.1 m)=0.25 m/s^2$

b2) Radial acceleration

The radial acceleration (also called centripetal acceleration) is given by:

$a_c =\omega^2 r$

where $\omega=25 rad/s$ is the final angular speed while r=0.1 m is the distance from the center of the wheel. Substituting numbers, we get

$a_c=\omega^2 r=(25 rad/s)^2(0.1 m)=62.5 m/s^2$

c) Number of revolutions

First of all, we need to find the angle covered during this time interval, which is given by:

$\theta=\omega_i t+ \frac{1}{2}\alpha t^2=\frac{1}{2} \alpha t^2=\frac{1}{2}(2.5 rad/s^2)(10 s)^2=125 rad$

And keeping in mind that $1 rev=2 \pi rad$, the number of revolutions made is:

$\theta = \frac{125 rad}{2 \pi rad/rev}=19.9 rev$

d) Deceleration

In this last part of the problem, we are told that the wheel comes to a stop after $\theta=5 rev=31.4 rad$. We also know the initial angular speed, $\omega_i =25 rad/s$, and the final angular speed, $\omega_f=0$, so we can find the new angular (de)celeration by using the equation:

$2\alpha \theta=\omega_f^2-\omega_i^2$

Substituting numbers, we get

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{0-(25 rad/s)^2}{2(31.4 rad)}=-9.95 rad/s^2$