19. [1pt] A bicycle wheel is mounted on a fixed, frictionless axle as shown. A m

Question

19. [1pt]
A bicycle wheel is mounted on a fixed, frictionless axle as shown. A massless string is wound around the rim and a constant horizontal force of magnitude F = 14 N starts pulling the string from the top of the wheel at time t = 0. The string does not slip on the wheel. The wheel has radius r = 0.210 m, mass m = 1.60 kg, and moment of inertia I = kmr2, where k = 0.800. Answer True, False or Cannot tell to the following statements; e.g., if the first is true, and the second is false, and one cannot tell for the third, enter TFC.You only have 3 tries.

i. The application of the force F causes to increase, regardless of the initial value of at t = 0.

ii. The work done on the wheel by the force is equal to the torque on the wheel.

iii. The magnitude of the angular acceleration of the wheel for t > 0 is given by = F/(kmr).

20. [1pt]
Assume the wheel is at rest at t = 0. What is the work done by the force F if the string is pulled through a distance d = 1.10 m, beginning at time t = 0?

21. [1pt]
Assuming the same conditions as in the previous problem, what is the magnitude of the angular velocity of the wheel after the string has been pulled through the distance d?

22. [1pt]
What is the angular displacement of the wheel after the string is pulled through the distance d?

23. [1pt]
How long does it take for the motion described in the previous three problems to occur?

Explanation / Answer

19, F = 14 N

r = 0.21 m

m = 1.6 kg

I = 0.8*mr^2 = 0.056448 kg m^2

i) True, as the applied torque gives anticlockwise angular acceleration, hence no mater what the w will always increase

ii) work done on the wheel by the force is not equal to the torque on the wheel, but it also includes other things like moment of inertial of the wheel etc

hence false

iii) I*alpha = F*r

hence

alpha = F*r/I = F*r/kmr^2 = F/kmr

hence true

hence

TFT

20. work done = F*R*theta

theta = 0.5*alpha*t^2

also, d = 1.1 m

hence work doen = F*d = 15.4 J

21. w = ?

0.5Iw^2 = 15.4

w = 23.3588 rad/s

22. theta = l/r = 5.238 radians

23. time = t

5.2380952380952380952380952380952 = 0.5*14/0.8*1.6*0.21 *t^2

t = 0.4484 s

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