Question

A motorcyclist is traveling along a road and accelerates for 4.50s to pass another cyclist. The angular acceleration of each wheelis +6.70 rad/s^2, and, just after passing, the angular velocity ofeach is +74.5 rad/s, where the plus signs indicate counterclockwisedirections. What is the angular displacement of each wheel duringthis time?

a. +221 rad
b. +131 rad
c. +335 rad
d. +355 rad
e. +267 rad

Answer 1

Angular displacement of the wheel,
`\theta=267.41\ rad`

Step-by-step explanation:

It is given that,

Angular acceleration of the wheel,
`\alpha =6.7\ rad/s^2`

Final speed of the wheel,
`\omega_f=74.5\ rad/s`

Time taken, t = 4.5 s

Initially, it is required to find the initial angular velocity of the wheel. Using the first equation of rotational kinematics as :

`\omega_f=\omega_o+\alpha t`

`\omega_o` is the initial speed of the wheel

`\omega_o=\omega_f-\alpha t`

`\omega_o=74.5-6.7* 4.5`

`\omega_o=44.35\ rad/s`

Let
`\theta` is the angular displacement of each wheel during this time. Using the second equation of motion as :

`\theta=\omega_o t+(1)/(2)\alpha t^2`

`\theta=44.35* 4.5+(1)/(2)* 6.7* (4.5)^2`

`\theta=267.41\ rad`

So, the angular displacement of each wheel during this time is 267 radian.