Question

Suppose an automobile engine can produce 215 N⋅m of torque, and assume this car is suspended so that the wheels can turn freely. Each wheel acts like a 13.5 kg disk that has a 0.155 m radius. The tires act like 2.05-kg rings that have inside radii of 0.18 m and outside radii of 0.32 m. The tread of each tire acts like a 9.5-kg hoop of radius 0.31 m. The 12.5-kg axle acts like a solid cylinder that has a 1.95-cm radius. The 31-kg drive shaft acts like a solid cylinder that has a 3.1-cm radius. 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:

α = 68.8 rad/s²

Explanation:

Given: m = 13.5 kg r= 0.155 m radius.  

Inertia is given by I = ½mr²

I = ½*13.5kg*(0.155m)²

I = 0.162 kg*m²

Given: m = 2.05kg ring inside radius ri= 0.18 m; outside radius ro= 0.32 m

I = ½m*(r₁² + r₂²)  

I = ½*2.05kg*(0.18² + 0.32²)

I = 0.138 kg*m²

Tread: m = 9.5 kg r= 0.31 m

I = mr² (hoop/ring or cylindrical shell formula)

I = 9.5kg*0.31m²

I = 0.9129kg*m²

Drive shaft: m= 31 kg solid cylinder r = 3.1 cm = 0.031m.

I = ½mr²

I = 31kg*0.031m²

I = 0.0298 kg*m²

Total moment of intertia

I = 0.162 + 0.1.38 + 0.9129 + 0.0298

It = 1.2427 kg*m²

τ = α*2*I ,τ = 95%*180N= 171N

α = τ/2*I

α = 171N/2*1.2427

α = 68.8 rad/s²