Question

4) A 1 meter-long, funny-shaped meter stick has an unknown mass but a center of gravity located at 0.75 m from the left end. When a metal wedge is placed at the middle of the bar and a .50-kg mass is hung from the left end, the bar balances perfectly on the metal wedge. a) Determine the mass of the funny-shaped meter stick (10 pts). b) Calculate the amount of upward force provided by the metal wedge (10 pts). 5) To admit water into the cooling unit of a nuclear reactor, you must rotate a very heavy disk (i =1/2 MR', R = 0.5 m, M = 300 kg through an angle of tradians is 5 seconds, otherwise...meltdown city. What angular acceleration is needed to cover the necessary angular displacement, assuming that the disk is starting from rest (10 points)? What is the force needed to produce this acceleration (assume you are pushing at right angle with the edge of the disk) (10 points)?

Answer 1

SOLUTION

PART (4):

(a)

In order to balance the meter stick, Anti-clockwise torques ($\tau_{1}$) should be equal to clockwise torques ($\tau_{2}$) .

0.25 m 0.5 m mg 0.5g

$\tau_{1}=Mg.L_{1}=0.5\;kg\times 9.8\;m/s^{2}\times0.5\;m=2.45\;N.m$

Let m be the mass of the meter stick, and given that the center of mass is at 0.75 m. Hence, distance to center of mass from axis/metal wedge is 0.25 m.

m x 9.8 m/s2 x 0.25 m 2.45m T2 mg.L2

We have:

$\tau_{1}=\tau_{2}$

$\therefore 2.45=2.45m$

$\therefore m=\frac{2.45}{2.45}\;kg$

$\boldsymbol{\therefore m=1\;kg}$

(b):

Normal force provided by the upward wedge:

$F=mg+Mg=(m+M)g$

$\therefore F=(1\;kg+0.5\;kg)\times9.8\;m/s^{2}$

$\boldsymbol{\therefore F=14.7\;N}$

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