1. A block of mass 9kg is moving in a circular path on a tabletop. The radius of

Question

1. A block of mass 9kg is moving in a circular path on a tabletop. The radius of the circle is 0.4m and the object's speed is 3m/s.

a) What is the initial angular momentum of the system? kg m^2/s

b) What is the initial kinetic energy of the system? J

Suppose the mass was being pulled in circular motion by a string. The string is threaded through a small hole on the top of the table, and a person pulls on the string until it has a length of 0.3m.

c) What is the new angular momentum of the system? kg m^2/s

d) What is the new speed of the mass? m/s

e) What is the new kinetic energy of the system? J

f) Comparing kinetic energies, how much work was done? J

2. A wheel with a radius of 0.6m and a moment of inertia 9kg m2 has an initial rotational velocity of 3 rad/s. A person takes a block and presses it against the edge of wheel, slowing it down with friction. The coefficient of kinetic friction between the wheel and the block is 0.3 and the person presses with a force that increases as a function of time: (7t) N.

a) How long does it take for the wheel to stop?  sec.
b) How far did the wheel rotate in the time it took to stop?  radians.

3. An clown is spinning with an angular momentum of 40 kg m2/s2. A torque of (4 + 2t)Nm acts upon the clown for 5 seconds.

What is the clown's final angular momentum?  kg m2/s2.

Explanation / Answer

Q1

a. J=mvr = 9*3*0.4 = 10.8 kg m^2/s

b. K=0.5*m*v^2 = 40.5 J

conservation of angular momentum(solved d first)

d. 10.8=9*v*0.3

v=4 m/s

c. J=mvr = 9*4*0.3= 10.8 kg m^2/s

e. K=0.5*m*v^2 = 72 J

f. W=delta(k) = 72-40.5=31.5 J

Q2.

a.moment of ineria of wheel= m*r^2 => 9=m*0.62

m=25 kg

v=rw = 0.6*3 = 1.8 m/s

negative acceleration = friction force/mass = uN/m = 0.3*7t/25

-a=-dv/dt

integration of (-adt) from 0 to t = integration of (-dv) from 1.8 to 0

0.084t2/2 = 1.8

t=6.54 sec

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