Question

Suppose an automobile engine can produce 180 N*m of torque, and assume this car is suspended so that the wheels can turn freely. Each wheel acts like a 15.5 kg disk that has a 0.175 m radius. The tires act like 1.9-kg rings that have inside radii of 0.19 m and outside radii of 0.315 m. The tread of each tire acts like a 12-kg hoop of radius 0.335 m. The 14.5-kg axle acts like a solid cylinder that has a 1.95-cm radius. The 32.5-kg drive shaft acts like a solid cylinder that has a 2.9-cm radius.
(a) calculate the angular acceleration in radians per square second, produced by the motor if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car.

Answer 1

Answer:

Explanation:

Moment of inertia of each wheel = 1/2 m R²

m is mass and R is radius of wheel

= .5 x 15.5 x .175²

= .2373 kg m²

moment of inertia of tyre

1/2 m ( r₁² + r₂² )

= 1/2 x 1.9 x ( .315² + .19²)

= 1/2 x 1.9 x ( .099+ .036)

= .12825 kg m²

moment of inertia of tread

= 1/2 m r²

= .5 x 12 x .335²

= .67335 kg m²

moment of inertia of axle

= 1/2 m r ²

= .5 x 14.5 x .0195²

= .00275

moment of inertia of drive shaft

= 1/2 x 32.5 x .029²

= .0137 kg m ²

Total moment of inertia of one tyre

=  1.05535 kg m²

total moment of inertia of two rear wheels

= 2.1107 kg m²

95 % of torque

= .95 x 180

= 171 Nm

angular acceleration

= torque / moment of inertia

= 171 / 2.1107

= 81.01 radian /s²