Stuck on physics rotational motion question.

Question

The tires of a car makes 65 revolutions as the car reduces its speed uniformly from 80km/h to 30km/h. The tires have a diameter of 8.0m. What was the angular acceleration of the tires?

 

I know the formulas but I can't seem to find the acceleration. Thank you in advance.

Answer 1

My first observation about this problem is that a tire diameter of 8.0m is unrealistically large.   Perhaps on a giant mining machine - but not on a car.   Another issue is the non SI unit of km/hr.  This must be changed to m/s.

80 km/hr = 22.2 m/s  and 30 km/hr = 8.33 m/s. 

With these preliminaries out of the way, the first thing to do is to convert the 65 revolutions to distance.

The distance (D) will be 65 pi 8 ( number of revolutions times circumference) =  1633 m (this is large because the tire diameter is large)

 

Since time is neither mentioned nor asked for, the v squared formula is the way to go

 

vf² =  vi² + 2 a D               this gives a way to calculate the linear acceleration (actually deceleration)

 

Plugging in vi = 22.2 , vf = 8.33 and D = 1633 gives   a =  -.130 m/s²

 

The angular acceleration, alpha, is   alpha = a/radius =   -  .0324  rad/s²

 

It is negative because it is a deceleration.

 

Answer 2

the answer I get is .032 rad / second² but I'm not 100% sure I did the math correctly.

 

the distance traveled during the deceleration is the circumference of the tires 2pi(.004km) x 65 = 1.6334 km

 

using the equation vf^2 - vi^2 = 2as, the linear deceleration is 1683.58 km / hr^2

 

from this you find the time of the deceleration which is .0296 hr = 107 seconds

 

Now use formula    delta v / delta t = radius (α)

 

50 km / hr = .004 km (α)

   107s

 

50 km                        = α

107s(3600s)(.004km)

 

= 0.032 rad / s^2