Question

A bicycle wheel is rotating at 41 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration of 0.49 rad/s. Part A What is the whoel's angular velocity, in rpm, 9.0 s later? Express your answer to two significant figures and include the appropriate units. w- Value Units Submit Request Answer Part B How many revolutions does the wheel make during this time? Express your answer to two significant figures and include the appropriate units. AXO RO? N of revs Submit Request Answer

Answer 1

A)

in 9 sec, it increases the angular velocity by (0.49 x 11) rad/s

= + 5.39 rad/s

now divide 5.39 by 2(pi) to give how many revolutions 5.17rad is equivalent to

= 0.8578...rps

multiply this by 60 to get it from rps to rpm increase

= +51.5 rpm

add this onto the starting revolution frequency of 41rpm to give:

92 rpm (rounded appropriately)...Answer.

B)

For the next part, find the average angular velocity (during acceleration) above the initial of 41rpm by dividing the additional 51.5 by 2.

= 25.75...rpm

Add this to the initial of 41rpm, giving 66.75...rpm is the average angular speed during the acceleration.

multiply the average speed by (9/60) because that's how long the time is in terms of minutes

(9/60) x 71.6849...rpm

= 10..revolutions (rounded appropriately)...Answer.

Hope that helps)