I need help with these questions if you could show your steps and work through t
ID: 2264525 • Letter: I
Question
I need help with these questions if you could show your steps and work through that would be awesome
A potter's wheel having a radius of 0.500 m and mass 45.0 kg is rotating feely at 50.0 rev/min clockwise direction. The potter can stop the wheel by pressing a wet rag against the outside rim of the wheel and exerting a radially inword force of 70.0 N. Since this force is in the radial direction, it alone cannot cause the wheel to slow its spinning. What can happen, however, is that a normal force can result from this inword force which results in a constant kinetic friction force. The friction force produces a constant torque that, in turn, causes the wheel to slow down. Assume that the wheel is a disk. Compute its rotational intertia. Suppose that after the wet rag is applied to the rim, it takes 6.00 seconds for the wheel to come to stop. What is the angular acceleration of the wheel? What net torque was exerted on the wheel? What normal force was exerted? What was the coefficient of kinetic friction between the wheel and the rag? A pulley is constructed of two cylinders as shown in the picture below. The bigger cylinder has a mass of 2.00 kg and a radius of 8.00 cm. The smaller cylinder has a mass of 1.50 kg and a radius of 60.0 cm. Find the rotational interia (moment of intertia) of the pulley if it is rotating about an axis through the centers of the two cylinders. Suppose two ropes are hung on this pulley: one rope is wrapped around the the outside cylinder and the second rope is wrapped around the inside cylinder. The ropes, through their tensions, produce forces acting on the cylinder as shown. Remember, the ropes come off a pulley at right angles to the radius. Find the angular acceleration of the pulley.Explanation / Answer
5)
a) MOI I = 0.5*M*R^2 = 0.5*45*0.5^2 = 5.625 Kg m^2
b) W1 = 50 *2*3.14/60 = 5.233 rad/s
alfa = W2-W1/t = 0-5.233 / 6 = -0.872 rad/s^2
c) torque Q = I*alfa = 5.625*0.872 = 4.905 Nm
d) Normal force, N = F = 70 N
e) Q = f*R
f = Q/R = 4.905 / 0.5 = 9.81
f = mue*N = mue*F
mue = f/N = 9.81/ 70=
N = 0.14
6) a) MOI = I = 0.5*1.50*0.6^2 + 0.5*2*0.8^2 = 0.91 kg m^2
b) torque Q1 = 80*0.8 = -64 Nm clock wise
torque Q2 = 60*0.6 = +36 Nm anticlock wise
Net torque = -64 + 36 = -28 Nm clock wise
Qnet = I*alfa
alfa = Qnet / I = 28/0.91 = 30.769 rad/s^2