Question

The tires of a car make 63 revolutions as the car reduces its speed uniformly from 91.0 km/h to 63.0 km/h. The tires have a diameter of 0.88 m.what wS the angular acceleration of the tires?

Answer 1

Angular acceleration,
`\alpha =-2.168\ rad/s^2`

Step-by-step explanation:

Given that,

Number of revolution,
`\theta=63\ rev=395.84\ rad/s`

Diameter of the car, d = 0.88 m

Radius, r = 0.44 m

Initial speed of the car, u = 91 km/h = 25.27 m/s

Initial angular speed,

`\omega_i=(u)/(r)`

`\omega_i=(25.27)/(0.44)`

`\omega_i=57.43\ rad/s`

Final speed of the car, v = 63 km/h = 17.5 m/s

Final angular speed,

`\omega_f=(v)/(r)`

`\omega_f=(17.5)/(0.44)`

`\omega_f=39.77\ rad/s`

To find,

The angular acceleration of the tires.

Solution,

The angular acceleration of the wire can be calculated using third equation of kinematics as :

`\alpha =(\omega_f^2-\omega_i^2)/(2\theta)`

`\alpha =(39.77^2-57.43^2)/(2(395.84))`

`\alpha =-2.168\ rad/s^2`

So, the angular acceleration of the tires is
`-2.168\ rad/s^2`.