A cord is wrapped around the rim of a solid uniform wheel 0.220 m in radius and

Question

A cord is wrapped around the rim of a solid uniform wheel 0.220 m in radius and of mass 7.40 kg . A steady horizontal pull of 42.0 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.

1) Compute the angular acceleration of the wheel.

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

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

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

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

Explanation / Answer

1) Torque due to the force, T = F*R

Moment of of inertia of wheel, I = 0.5*M*R^2

Let alfa is teh anularar acceleration.

Now Apply, T = I*alfa

==> alfa = T/I

= F*R/(0.5*M*R^2)

= F/(0.5*M*R)

= 42/(0.5*7.4*0.22)

= 51.6 rad/s^2

2) a_tan = R*alfa

= 0.22*51.6

= 11.35 m/s^2

3) net force on wheel = 0

F - F_axle = 0

F_axle = F

= 42 N

4) towards left

5) D) would change

that means the direction of force exerted by axle on the will changes.

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