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Pavel [41]
3 years ago
13

A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t

o the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. What is the wheels angular acceleration? What force does the axle apply to the wheel? What torque does the axle apply to the wheel?
Physics
1 answer:
Grace [21]3 years ago
3 0

Answer:25.97 rad/s^2

Explanation:

Given

radius of wheel r=0.28 m

mass of wheel m=8.80 kg

Force F=32 N

Moment of Inertia of solid wheel I=\frac{mr^2}{2}

I=\frac{8.8\times 0.28^2}{2}

I=0.344 kg-m^2

Torque is given by

\tau =F\times r=I\times \alpha

32\times 0.28=0.344\times \alpha

\alpha =25.97 rad/s^2

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

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