A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and

Question

A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 7.20 kg. A steady horizontal pull of  32.0 N to 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.

Part A

Compute the angular acceleration of the wheel.

Part B

Compute the acceleration of the part of the cord that has already been pulled off the wheel.

Part C

Find the magnitude of the force that the axle exerts on the wheel.

Part D

Find the direction of the force that the axle exerts on the wheel.

Part E

Which of the answers in parts (A), (B), (C) and (D) would change if the pull were upward instead of horizontal?

Explanation / Answer

Given that

A cord is wrapped around the rim of a solid uniform wheel of radius (r) =0.300 m

And of mass (m) =7.20 kg.

A steady horizontal pull of (F) = 32.0 N to the right

We know that

Torque =I*alpha

F*r =(1/2)mr2*alpha

a)

Then the angular acceleration of the wheel is alpha =2*F/mr =2*32/(7.20kg)(0.300m) =29.63rad/s2

b)

The acceleration of the part of the cord that has already been pulled off the wheel is given by

a =alpha*r =(29.63)(0.300) =8.889m/s2

c)

The magnitude of the force that the axle exerts on the wheel is given by

F =-32Nx+(7.20kg)(9.81m/s2)y =-32Nx+70.632Ny

Then the magnitude of force is given by

F =Sqrt[(32)2+(70.632)2] =Sqrt[1024+4988.879] =77.542N

d)

The direction of the force that the axle exerts on the wheel is given by

Theta =tan-1(Fy/Fx) =tan-1(70.632/-32) =-65.626degrees

Then the direction of force is given by theta =180-65.626 =114.373degrees

E)

Alpha will be remain unchanged

Aacceleration will change, because gravity will oppose the applied force 32N

The magnitude of force will also be less because the gravity will not oppose the applied force.The answer c and d will change only because there will be only the vertical force

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