In the figure, block 1 has mass m 1 = 430 g, block 2 has mass m 2 = 510 g, and t

Question

In the figure, block 1 has mass m1 = 430 g, block 2 has mass m2 = 510 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.7 cm. When released from rest, block 2 falls 79 cm in 4.9 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answersalong the way to the solution. Use g = 9.81 m /s2

Explanation / Answer

T1 - m1*g = m1*a

or T1 = m1*g + m1*a

sum forces on m2

m2*g - T2 = m2*a

or T2 = m2*g - m2*a

sum torques about the pulley

T2*r - T1*r = I* = I*a/r

or (m2*g - m2*a)*r - (m1*g + m1*a)*r = I*a/r

but a is determined from the kinematic data

a. ) applying netwons laws of motion

y = 1/2*a*t^2

so, a = 2*y/t^2

= 2*0.79m/4.9^2

a= 5.1x10^-2 m/s^2

b.) put the value of a in above equation

T1 = m1*g + m1*a

= 0.43*9.81 + 0.43 * 5.1 x 10 ^-2

=4.196 N

C.) similarly

T2 = m2*g - m2*a

=0.51*9.81 - .51* 5.1 * 10^-2

=4.9 N

D.) angular acceleration = a/r

= 5.1 * 10 ^-2 .0.47

=2.297 * 10^-2 rad/sec^2

E.)so I = r/a*((m2*g - m2*a)*r - (m1*g + m1*a)*r)

I = 0.047/5.1x10^-2*((0.510*9.8 - 0.510*5.1x10^-2)*0.0470 -(0.430*9.8 + 0.430*5.1x10^-2)*0.0470)

= 0.921 ((.233) - (.197 ) kg-m^2

moment of inertia (I) = 0.0333 kg-m^2

=2.14 * 10 ^-5

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