A classic Atwood machine has a small mass of 10kg and a large mass of 10.4kg. Th

Question

A classic Atwood machine has a small mass of 10kg and a large mass of 10.4kg. The pulley is a 1.7kg hoop with a diameter of 54cm. If the friction in the pulley is very small, find the tension in each of the ropes while the system is in motion. o Consider an old time barbell. It consist of two, 6cm radius solid spheres each with a mass of 9kg connected by a 2kg cylinder that has a diameter of 2cm and is 10 cm long. Find its moment of inertia about the center of mass. Find the moment of inertia about the center of the right sphere as shown. o I have a bucket filled with water. The bucket is 10cm high with a diameter of 8cm. There is a 76cm rope tied to the top of the bucket. How quickly must I swing this bucket in a vertical circle such that there is no spillage when the bucket is upside down? o Design, for me, a road whose curvature is 175m. I want the speed limit to be about 45 mph. I have a set of point masses located at the given locations: o m1-18kg @(1,3.3)m m3 = 0.9kg @(5.7,-7.2)m m2 15.5kg @(2.4,0)m m4-9.3kg @(0,-2.5)m ms = 1.4kg @(-3.1,-4.3)m Find the center of mass and the moment of interia about each axis.

Explanation / Answer

1)Given,

m1 = 10 kg ; m2 = 10.4 kg ; M = 1.7 kg ; r = 54 cm

Let T1 and T2 be the tensions and a be the acceleration

Sum of forces in Y direction:

T2 - m2g = - m2 a

T2 = m2 (g - a)

T1 = m1 (g + a)

T1R - T2R = -I(cm) alpha

a = R alpha => alpha = a/R

putting the values of alpha, T1 and T2 we get

m1 (g + a)R - m2 (g - a) R = 1/2 M R^2 x a/R

solving for a we get

a = m2g - m1g/(m1 + m2 + M/2)

a = 9.81(10.4 - 10)/(10 + 10.4 + 1.7) = 0.178 m/s^2

T1 = 10 x (9.81 - 0.178) = 96.32 N

T2 = 10.4 (9.81 + 0.178) = 103.88 N

Hence, T1 = 96.32 and T2 = 103.88 N

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