Question

An automobile moves on a road with a speed of 54 km/h. The radius of its wheels is 0.35 m. What is the average negative torque transmitted by its brakes to a wheel if the vehicle is brought to rest in 15s? Moment of inertia of wheel about axis of rotation is 3 kgm^2. If the torque is due to a force applied tangentially on the rim of the vehicle, what is the magnitude of the force?\

Answer 1

HEY DEAR ...

Given Conditions ⇒

Initial linear velocity(u) = 54 km/hr.
 = 54 × 5/18 [Changing into m/s.]
 = 15 m/s.
Radius of the wheel(r) = 0.35 m.

Now, Using the Formula,

Initial Angular Velocity (ω₁) = u/r
ω₁ = 15/0.35
ω₁ = 42.86 radian/sec.

Also, Final Angular Velocity (ω₂) = 0

Now, Using the Formula,
Angular Acceleration (α) = (ω₂ - ω₁)/t
 = (0 - 42.86)/15
 = - 2.86 radian/sec²

Moment of Inertia about the axis of the rotation(I) = 3 kg-m²

∴ Torque = I × α
⇒ Torque = 3 × -2.86
⇒ Torque = -8.58 Nm.


Hence, the average negative torque is -8.58 N-m.

HOPE , IT HELPS ...