Question

A car accelerates uniformly from rest and reaches a speed of 21.9 m/s in 9.01 s. Assume the diameter of a tire is 59.0 cm.(a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. ... rev(b) What is the final angular speed of a tire in revolutions per second? ...rev/s

Answer 1

To solve this problem we will apply the linear motion kinematic equations. There we will find the acceleration, which will allow us to find the total displacement. From the relationship given between the circumference of a circle and the total length, we can determine the number of revolutions.

For the second part we will use the relationship between angular and linear velocities given through the radius of the circumference to calculate the final value.

PART A) Calculate the acceleration of the car

`v= v_0+at`

Where,

a = Acceleration

`v_0` = Initial velocity

v = Final velocity

t = time

Replacing we have that

`21.9=0+a*9.01s`

`a = 2.43m/s^2`

Calculate the displacement

`x = v_0t+(1)/(2) at^2`

Replacing

`x = 0*9.01 + (1)/(2) (2.43)(9.01)`

`x = 98.65m`

Calculate the circumference of wheel:

`\phi = \pi d`

Where,

d = Diameter

`\phi` = Circumference

`\phi = \pi (59*10^(-2))`

`\phi = 1.85m^2`

Calculate the number of revolutions

`N = (x)/(\phi)`

`N = (98.65m)/(1.85m^2)`

`N = 53.32rev`

Therefore the number of revolutions is 53.32rev

PART B )

Calculate the angular velocity

`v = r\omega`

`\omega = (v)/(r) \rightarrow r = (d)/(2)`

`\omega = (21.9)/(29.5*10^(-2))`

`\omega = 74.23rad/s`

Therefore the final angular speed is 74.23rad/s