[A] A uniform density meter stick is pivoted at the 20 cm mark. The mass of the

Question

[A] A uniform density meter stick is pivoted at the 20 cm mark. The mass of the meter stick is 250 grams. A 750g mass is to be hung from the meter stick at x to balance the whole assembly. Find x.

[B] A pulley is made from a solid uniform disc with a 5cm radius and whose mass = 200g. Attached to the outer rim is a thin ring (black) with a mass of 200g (treat the ring as if all its mass was at r = 5cm).

(1) Calculate the moment of inertia for this pulley.

(2) If a string wrapped around this pulley is attached to a 2kg mass, calculate the torque on the

pulley and the linear acceleration of the mass if the string pulls off the pulley without slipping.

(3) If this acceleration lasted for 3 seconds, starting from rest, determine the final angular

velocity and the total angle the pulley has rotated through.

[C] A mass M=3kg is hung from a rod of mass m=1kg, 3/4 of the way to the end. There is a horizontal wire attached to the end of the rod pulling on it with tension Ft . The bottom of the rod is hinged to the wall making an angle of 60 degrees with the wall. The assembly is to remain at rest.

(1) Set up the linear force balance equations, one along each axis you choose.

(2) Choose a pivot point and set up the torque balance about that point.

(3) Solve for the tension in the wire and the two force components at the hinge.

Note: I will only credit and rate if you show all of your work.

Explanation / Answer

A)
250*20 = 750*d

d = 250*20/750

= 6.67 cm from pivot

from end, x = 20-6.67 = 13.33 cm

B)

1)I = 0.5*m*r^2 + m*r^2 = 0.5*0.2*0.05^2 + 0.2*0.05^2 = 7.5*10^-4 kg.m^2

2) T = r*M*g

= 0.05*2*9.8

= 0.98 N

3) alfa = T/I

= 0.98/7.5*10^-4

= 1306.7 rad/s^2

w = alfa*t = 1306.7*3 = 3920 rad/s



C)

1)let fx and fy are forces are reaction forces at hinge

Fnet,x = 0

Ft - fx = 0



Fnety = 0

fy - m*g - M*g = 0


2) m*g*(L/2)*sin(60) + M*g*(3*L/4)*sin(60) - Ft*L*sin(30) = 0


3)

Ft = (m*g*0.5*sin(60) + M*g*0.75*sin(60))/sin(30)

= (1*9.8*0.5*sin(60) + 3*9.8*0.75*sin(60)) / sin(30)

= 46.68 N

now, fx = Ft = 46.68 N

fy = m*g + M*g

= 9.8+ 3*9.8

= 39.2 N

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