An electric motor turns a flywheel through a drive belt that joins a pulley on t

Question

An electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel as shown in the figure below. The flywheel is a solid disk with a mass of 53.5 kg and a radius R = 0.625 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of 0.230 m. The tension Tu in the upper (taut) segment of the belt is 167 N, and the flywheel has a clockwise angular acceleration of 1.67 rad/s2. Find the tension in the lower (slack) segment of the belt.
N

A block of mass m1 = 1.65 kg and a block of mass m2 = 6.20 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. The fixed, wedge-shaped ramp makes an angle of ? = 30.0° as shown in the figure. The coefficient of kinetic friction is 0.360 for both blocks.

(a) Draw force diagrams of both blocks and of the pulley.

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(b) Determine the acceleration of the two blocks. (Enter the magnitude of the acceleration.)
m/s2

(c) Determine the tensions in the string on both sides of the pulley.

The combination of an applied force and a friction force produces a constant total torque of 36.2 N · m on a wheel rotating about a fixed axis. The applied force acts for 5.90 s. During this time, the angular speed of the wheel increases from 0 to 9.8 rad/s. The applied force is then removed, and the wheel comes to rest in 59.5 s.

(a) Find the moment of inertia of the wheel.
kg · m2

(b) Find the magnitude of the torque due to friction.
N · m

(c) Find the total number of revolutions of the wheel during the entire interval of 65.4 s.
revolutions

Consider the system shown in the figure below with m1 = 30.0 kg, m2 = 14.4 kg, R = 0.230 m, and the mass of the pulley M = 5.00 kg. Object m2 is resting on the floor, and object m1is 4.30 m above the floor when it is released from rest. The pulley axis is frictionless. The cord is light, does not stretch, and does not slip on the pulley.

(a) Calculate the time interval required for m1 to hit the floor.
?t1 =  s

(b) How would your answer change if the pulley were massless?
?t2 =  s

Explanation / Answer

Q3.The combination of an applied force and a friction force produces a constant total torque ......

Part A

We know that

wf = wi + alpha*t

alpha = (wf - wi)/t

alpha = (9.8 - 0)/5.90 = 1.66 rad/sec^2

Now We know that

Net torque = I*alpha

I = Net torque/alpha

I = 36.2/1.66

I = 21.81 kg-m^2

Part B.

Again using

w0 = wf + alpha1*t1

alpha1 = (0 - 9.8)/59.5 = -0.165 rad/sec^2

Torque due to friction will be

Tf = I*alpha1 = 21.81*(-0.165) = -3.60 N-m

|Tf|= 3.60 N-m

Part C

Number of revolution, during first 5.90 sec

R1 = wi*t + 0.5*alpha*t^2

R1 = 0*5.90 + 0.5*1.66*5.90^2 = 28.89 rad

R1 = 28.89/(2*pi) = 4.60 rev

During next 59.5 sec

R2 = wi*t + 0.5*alpha1*t1^2

R2 = 9.8*59.5 + 0.5*(-0.165)*59.5^2 = 291.02 rad

R2 = 291.02/(2*pi) = 46.32 rev

Total revolutions = R = 4.60 + 46.32 = 50.92 rev

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