(i) The pulley mass is \"m \", what is the acceleration of the system? (ii) If m
ID: 2069661 • Letter: #
Question
(i) The pulley mass is "m", what is the acceleration of the system?
(ii) If m2 start h (meters) from the ground what is the speed when m2 hit the ground
(iii) you also can get that speed using work and energy equation. Please do it.
Blocks of mass m1 and m2 are connected by a massless string that passes over the pulley in the figure . The pulley turns on frictionless bearings, and mass m1 slides on a horizontal, frictionless surface. Mass m2 is released while the blocks are at rest. (i) The pulley mass is m, what is the acceleration of the system? (ii) If m2 start h (meters) from the ground what is the speed when m2 hit the ground (iii) you also can get that speed using work and energy equation. Please do it.Explanation / Answer
a) The net force accelerating m2 is = m2*g - T (with T = tension) m2g - T = m2a --> T = m2g - m2a. And the tension is also the force that accelerates m1: T = m1a set both equations equal to get m2g - m2a = m1a --> solve for a: a(m1 + m2) = m2g a = m2g/(m1+m2) T = m1*m2*g/(m1+m2) b) T1 is the tension between m1 and the pulley T2 is the tension between m2 and the pulley the net force on m1 is T1 = m1a the net force on m2 is m2g - T2 = m2a --> T2 = m2g - m2a the net force on the pulley is T2 - T1 and the torque ? on the pulley is then = F*R = (T2 - T1)R and the torque ? on the pulley is also ? = I*? = I*a/R (with I = moment of inertia of pulley, ? = angular acceleration of pulley) --> (T2-T1)R = I*a/R --> divide by R (T2 - T1) = Ia/R^2 plug in the expressions for T1 and T2: (m2g - m2a) - m1a = Ia/R^2 --> solve for a: a(-m2 - m1 - I/R^2) = - m2g a = m2g/(m1+m2+I/R^2) with I = 1/2*mp*R^2 ---- plug a into the two equations for T1 and T2: T1 = m1a T2 = m2g-m2a
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.