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 8.80 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.

a. Compute the angular acceleration of the wheel in rad/s^2.

b. Compute the acceleration of the part of the cord that has already been pulled off the wheel in m/s^2.

c. Find the magnitude of the force that the axle exerts on the wheel in N.

d. Find the direction of the force that the axle exerts on the wheel. (Degrees below the horizontal, away from the direction of the pull on the cord.)

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

(A), (B), (C) and (D)

(A) and (B)

(C) and (D)

(A) and (D)

(A), (B), (C) and (D)

(A) and (B)

(C) and (D)

(A) and (D)

Explanation / Answer

The angular acceleration equation is:

F*R = I*dw/dt ==> dw/dt = F*R/I

for a wheel/disc, I = 1/2MR^2, so dw/dt = 2F/(M*R) = 43.388 rad/s^2<--- ans A

the cord accelerates with a linear acceleration proportional to the angular acceleration of the wheel (assuming a negligible cord mass):

a-cord = dw/dt * R = 9.54 m/s^2.<---- ans B

the axle exerts an horizontal force which opposes F so no linear acceleration result.
there is also a vertical force to neutralize the weight (normal)

C. sqrt(42^2+8.8^2*9.81^2)
96 N

D) atan(8.8*9.81/42) below horizontal
64.06  degrees below horizontal

pulling upward:

ans A does not change.
ans B does not change
ans C and D change:

there is only a vertical axle force, Faxle-ver = Mg + F = 88 + 42 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.