Question

Part A A bicycle wheel is rotating at 41 rpm when the cyclist begins to pedal harder, glving the wheel a constant angular acceleration of 0.49 rad/s. What is the wheel's angular velocity, in rpm, 9.0 s later? Express your answer to two significant hures and include the appropriate units. w. 92 rpm Submit Previous Answers Request Answer X Incorrect; Try Again; 8 attempts remaining Part B How many revolutions does the wheel make during this time? Express your answer to two significant figures and include the appropriate units. IVO AXD Om ? N of rovs - 10 Submit Previous Answers Request Answer

Answer 1

Partb A)

Using equation of the motion

$\omega =\omega _{o}+\alpha t$ ...1

Here $\omega _{o}=41 rpm=41*2\pi /60=4.29 rad/s$ ,$\alpha =0.49 rad/s^{2}$ , t=9.0s

Put these value in equation 1

$\omega =4.29+0.49*9=8.7 rad/s$

since 1 rad=1/2$\pi$ rev, 1s=1/60 min so

$\omega =8.7 rad/s=8.7*(1/2\pi )*60=83.08 rpm$

$\omega =83.08 rpm$

=83 rpm Answer

part B)

Using equation of the motion

$\theta =\omega _{o}t+\alpha t^{2}/2$

$\theta =4.29*9+0.49*(9)^{2}/2$

$\theta =58.455 rad$

Since 1rad  =(1/2$\pi$ ) rev

So

$\theta =58.455 /2\pi =9.30 rev$

$\theta=9.30 rev$

=9.3 rev Answer