Question

A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a time interval of 5 s, what is the angle in radians through which the wheel rotates? rad(c) At t=10 s the angular speed begins to increase at a rate of 1.6 rad/s/s. At t=15 s, what is the angular speed in radians per second? rad/s(d) Through what angle in radians did the wheel rotate during the time between t=10 s and t=15 s? rad(e) If the wheel rolls along the ground without slipping, the instantaneous velocity of the atoms of the object that are momentarily in contact with the ground is zero. This zero-velocity condition implies that vCM=ω⁢R, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ω⁢R. During the time between t=10 s and t=15 s, how far did the center of the wheel move, in meters? m

Answer 1

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$