Calculating Angular Acceleration of Bicycle Wheels

[cristina]

A bicycle has wheels of 1.2m diameter. the bicyclist accelerates from rest with constant acceleration to 24km/h in 14.0s. What is the angular acceleration of the wheels?

If the bicycle is going forwards relative to the ground with a speed of 24km/h, then, all points on the tread are moving around the wheel with a speed (s) of 24km/h relative to the axel.

s = 24km/h = 6.7m/s (I know the significant figures rules but I don't know where to round)
r = 1.2m/2 = 0.6m

w=s/r
w=6.7/0.6 = 11.2 rad/s ( is my rounding correct here?)

a=v^2/r=(6.7)^2/0.6= 74.8m/s^2


Is the reasoning correct?

Thank you for your help.

 

[outandbeyond2004]

A number in the problem has three digits (12.0), so that is what your answer should have at most, also.

The number 6.7/0.6 ought to have just 2 figures, but since the first number really should have 3 figures in the first place, and the second number is accurate to 3 figures, . . .

Tangential acceleration = angular acceleration * radius
is another formula you could use to check your work, which I only eyeballed. Looks OK to me.

 

[ShawnD]

A bicycle has wheels of 1.2m diameter. the bicyclist accelerates from rest with constant acceleration to 24km/h in 14.0s. What is the angular acceleration of the wheels?

If the bicycle is going forwards relative to the ground with a speed of 24km/h, then, all points on the tread are moving around the wheel with a speed (s) of 24km/h relative to the axel.

s = 24km/h = 6.7m/s (I know the significant figures rules but I don't know where to round)
r = 1.2m/2 = 0.6m

This is corect

w=s/r
w=6.7/0.6 = 11.2 rad/s ( is my rounding correct here?)

This is correct

a=v^2/r=(6.7)^2/0.6= 74.8m/s^2

The question asks for angular acceleration.

a = (delta w)/t
a = (11.2)/(14)
a = 0.8 rad/s^2