Mass and Kinetic Energy

955 words 4 pages
Disk With Weight:

A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3 kg weight is hanging on the string. The system of the weight and disk is released from rest.

a) When the 3 kg weight is moving with a speed of 2.2 m/s, what is the kinetic energy of the entire system? KETOT = KEwheel+KEweight = (1/2)(I)(w2)+(1/2)(m*v2) =(0.5* v2)(m+1/2M) =0.5*(2.2^2)*(3+(.5*15)) J

b) If the system started from rest, how far has the weight fallen? H = KETOT/MG = 0.5*(2.2^2)*(3+(.5*15))/(3*9.8) m

c) What is the angular acceleration at this point? Remember that a = αR, or α = a/R

Solve for acceleration by using
…show more content…

What is the smallest value of the coefficient of friction µ such that the sign will remain in place?
= .282


A man of mass mm = 95 kg decides to paint his house. To do this, he builds a platform using a uniform beam with a mass of mb = 100 kg and a length of L = 7 meters. The beam is supported by two sawhorses, as shown in the diagram above.

a) If the man stands over the support at point B, calculate the force exerted by the beam on the support at A. 490

b) How far from the end of the beam (the end closest to support A) does the person have to stand to unbalance the beam? 6.57

c) Later that day, after thinking about how cool rotational dynamics really is, the man decides to conduct an experiment. He removes one of the supports and places the other as shown in the diagram. Standing at the end of the board, he has his daughter place paint cans, each of mass mc = 2.9375 kg, on the opposite end. How many cans will the girl have to place on the board to provide the best balance? (You may neglect the actual length of the board that both the man and the cans occupy. Assume both are points at the ends of the board.) Also, take the axis of rotation about the sawhorse.


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