Question 1: The masses are Q=0.700 kg, R=0.400 kg, and S=0.800 kg. Calculate the

Question

Question 1:

The masses are Q=0.700 kg, R=0.400 kg, and S=0.800 kg. Calculate the moment of inertia (of the 3 masses) with respect to an axis perpendicular to the xy plane and passing through x=0 and y=-2. [Since the masses are of small size, you can neglect the contribution due to moments of inertia about their centers of mass.]

Question 2:

A crate with a mass of 191.5 kg is suspended from the end of a uniform boom with a mass of 79.1 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.

Question 3:

A small mass M attached to a string slides in a circle (x) on a frictionless horizontal table, with the force F providing the necessary tension (see figure). The force is then increased slowly and then maintained constant when M travels around in circle (y). The radius of circle (x) is twice the radius of circle (y).

( true false greater than less than equal to) While going from x to y, there is no torque on M
( true false greater than less than equal to) M's kinetic energy at y is twice that at x.
( true false greater than less than equal to) As M moves from x to y, the work done by F is .... 0.
(true false greater than less than equal to ) M's angular momentum at y is .... that at x.
(true false greater than less than equal to) M's angular velocity at y is four times that at x.

Explanation / Answer

(b)

m1 = 0.7 kg        r1 = sqrt(5^2 + 2^2) = 5.4 m

m2 = 0.4 kg        r2 = sqrt(5^2+4^2) = 6.4 m

m3 = 0.8 kg        r3 = sqrt(3^2+1^2) = 3.16 m

moment of inertia = I = m1*r1^2 + m2*r2^2 + m3*r3^2

I = (0.7*5.4^2) + (0.4*6.4^2) + (0.8*3.16^2) = 44.78 kg m^2 <<-----answer

2)

m1 = 191.5 kg

w1 = m1*g

r1 = L

m2 = 79.1 kg

w2 = m2*g

r2 = L/2

from the figure

tan beta = 4/9 =========> beta = 24

tan alfa = 4/9 ===========> alfa = 24

net torque about the hinge = 0

T*cosbeta*L*sinalfa + T*sinbeta*L*cosalfa = w1*r1*cosalfa + w2*r2*cosalfa

T*L*sin(alfa+beta) = w1*r1*cosalfa + w2*r2*cosalfa

T*L*sin(24+24) = (191.5*9.8*L*cos24)+(70.1*9.8*L/2*cos24))

T*sin(24+24) = (191.5*9.8*cos24)+(70.1*9.8*(1/2)*cos24)

T = 2729.3 N <<---------answer

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