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# Problem #1 (20pts) Two weights of differing masses mi and m- (where m.docx

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# Problem #1 (20pts) Two weights of differing masses mi and m- (where m.docx

Problem #1 (20pts) Two weights of differing masses mi and m: (where m is larger than ms ) are connected string that passes over a frictionless pulley in the shape of a uniform cylinder of mass a moment of inertia 1-(1/2) M.R. There is no slipping between the string system is initially at rest. M, radius R, and the pulley. The a) Find a beautiful algebraic expression for the magnitude of the lincar spsed v of mi afher having descended a distance h from the starting point. Neglest air dnug & E M with b) Find the an al?ebraic espression for the lisar asslanation a of tho system 2. for each one of the tonsions. Ti and T, in each cable. c) Find an algebraic expression initial: MR n2 Page 2 of 2
Solution
On the right side of the pulley,
m1g - T1 = m1*a
T2 - m2g = m2*a
(T1 - T2)*R = 0.5*M*R^2*alpha
and R*alpha = a
So (T1 -T2) = 0.5*M*a
So Adding the first and second equation gives,
(m1 - m2)*g -(T1 -T2) = (m1 +m2)*a
So a = (m1 -m2)*g/(m1 + m2 +M/2)
v^2 = 2*a*h = 2*h*(m1 -m2)*g/(m1 + m2 +M/2)
v = sqrt(2*h*(m1 -m2)*g/(m1 + m2 +M/2))
T1 = m1*(g-a) = m1*g*(1 - (m1-m2)/(m1 + m2 +M/2))
T2 = m2*(g+a) = m2*g(1+ (m1-m2)/(m1 + m2 +M/2))
.

Problem #1 (20pts) Two weights of differing masses mi and m: (where m is larger than ms ) are connected string that passes over a frictionless pulley in the shape of a uniform cylinder of mass a moment of inertia 1-(1/2) M.R. There is no slipping between the string system is initially at rest. M, radius R, and the pulley. The a) Find a beautiful algebraic expression for the magnitude of the lincar spsed v of mi afher having descended a distance h from the starting point. Neglest air dnug & E M with b) Find the an al?ebraic espression for the lisar asslanation a of tho system 2. for each one of the tonsions. Ti and T, in each cable. c) Find an algebraic expression initial: MR n2 Page 2 of 2
Solution
On the right side of the pulley,
m1g - T1 = m1*a
T2 - m2g = m2*a
(T1 - T2)*R = 0.5*M*R^2*alpha
and R*alpha = a
So (T1 -T2) = 0.5*M*a
So Adding the first and second equation gives,
(m1 - m2)*g -(T1 -T2) = (m1 +m2)*a
So a = (m1 -m2)*g/(m1 + m2 +M/2)
v^2 = 2*a*h = 2*h*(m1 -m2)*g/(m1 + m2 +M/2)
v = sqrt(2*h*(m1 -m2)*g/(m1 + m2 +M/2))
T1 = m1*(g-a) = m1*g*(1 - (m1-m2)/(m1 + m2 +M/2))
T2 = m2*(g+a) = m2*g(1+ (m1-m2)/(m1 + m2 +M/2))
.

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### Problem #1 (20pts) Two weights of differing masses mi and m- (where m.docx

1. 1. Problem #1 (20pts) Two weights of differing masses mi and m: (where m is larger than ms ) are connected string that passes over a frictionless pulley in the shape of a uniform cylinder of mass a moment of inertia 1-(1/2) M.R. There is no slipping between the string system is initially at rest. M, radius R, and the pulley. The a) Find a beautiful algebraic expression for the magnitude of the lincar spsed v of mi afher having descended a distance h from the starting point. Neglest air dnug & E M with b) Find the an al?ebraic espression for the lisar asslanation a of tho system 2. for each one of the tonsions. Ti and T, in each cable. c) Find an algebraic expression initial: MR n2 Page 2 of 2 Solution On the right side of the pulley, m1g - T1 = m1*a T2 - m2g = m2*a (T1 - T2)*R = 0.5*M*R^2*alpha and R*alpha = a So (T1 -T2) = 0.5*M*a So Adding the first and second equation gives, (m1 - m2)*g -(T1 -T2) = (m1 +m2)*a So a = (m1 -m2)*g/(m1 + m2 +M/2) v^2 = 2*a*h = 2*h*(m1 -m2)*g/(m1 + m2 +M/2) v = sqrt(2*h*(m1 -m2)*g/(m1 + m2 +M/2)) T1 = m1*(g-a) = m1*g*(1 - (m1-m2)/(m1 + m2 +M/2)) T2 = m2*(g+a) = m2*g(1+ (m1-m2)/(m1 + m2 +M/2))