Question

An car traveling at 110.0 km/h has tires of 67.0 cm diameter. If the car is brought to a stop uniformly in 45.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels?

Answer 1

u (speed) = 110km/hr

= 110 * 5/18 m/s

= 30.55 m/s

Diameter = 67 cm

Radius = 67/2 = 33.5 cm

= 0.335 m

No. of turns (N) = 45

V ( final speed) = 0 m/s (Rest)

Using, equation of motion

w^2 - W_o^2 = 2

where w is the final angular velocity

w_o = initial angular velocity

is angular acceleration

angular displacement

we know that, V = R w

Substituting the given values in the above formula

0 - (30.55/0.335)^2 = 2** (45*2)

-8316.35 = 565.58

= - 14.7 rad/sec^2

Magnitude of angular acceleration of the wheels = 14.7 rad/sec^2

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