A man holding a heavy object in each hand stands on a small platform that is fre

Question

1. A man holding a heavy object in each hand stands on a small platform that is free to rotate about a vertical axis. Initially he is standing with his arms outstretched and he and the platform are rotating with an angular velocity of 0.800 rad/s. With his arm outstretched, the moment of inertia of the system (man + platform + weights) is 4.00 kg.m^2. Then he pulls the weights in close to his chest, and the moment of inertia of the system becomes 3.00 Kg.m^2.   a. What is the angular velocity of the man and platform after he has pulled in the weights? b. How much work did the man do when he pulled his arms in?   
2. A large wheel with radius R= 0.200 m is mounted on a frictionless axle that passes through the center of the wheel. A light rope is wrapped around the wheel and a block of mass 5.00 Kg is suspended from the free end of the rope. The system is released from rest and the block moves downward. As the block is descending, the tension in the rope is 36.0 N. a. As the block descends, what is the angular acceleration of the wheel? b. What is the moment of inertia of the wheel, for an axis at its center?   
3. A block with mass m= 0.300 kg is attaced to one end of a horizontal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x= +0.240 m, its acceleration is ax = -12.0 m/s^2 and its velocity is Vx = +4.00 m/s. a. What is the force constant k of the spring? b. What is the amplitude of the motion? c. What is the maximum speed of the block during its motion? d. What is the maximum acceleration of the block during its motion?

Explanation / Answer

1) momentum of system is conserved so,

4*0.8 = 3*w ... were w is angular momentum

w = 3.2/3 = 1.0667 rad/sec

work done = 0.5 * ( 3 * (3.2 / 3)^2 - 4 * (0.8)^2 ) = 0.4267 J

2)

a) I = m * ( r^2 )/ 2= 5 * ( (0.2)^2) / 2 = 0.1kg*m^2 ...were I is moment of inertia of wheel about x axis and a is angular acceleration

=> I * a = 36 * 0.2

so a = 72 rad / sec^2

b) I = m * ( r^2 ) = 0.2 kg*m^2

3)

a)-K*x = m*a

-K*0.24 = 0.3*(-12)

Hence K = 15 N/m

b) 1/2 * ( K*x^2 + m*V^2) = 1/2 * K *A^2

were A is amplitude = 0.6145 m

c) A*w = V

w = (15/0.3)^0.5 = 7.0711 sec^(-1)

so V = 7.0711*0.6145 = 4.3451m/sec

d) K*A = m*a

15*0.6145 = 0.3*a

so maximum a = 30.725 m/sec^2

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