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iew History Bookmarks Window Help edigen of Physics E the figure he... Google Acco. Q Search Done ice Gradebook ORION Downloadable eTextbook ment HESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSTON 4BACK NEXT In the figure, block 1 has mass m 478 g, block 2 has mass m2-571 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.62 cm. When released from rest, block 2 falls 76.0 cm in 5.06 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 (e) tension Ti (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 answers along the way to the solution. Use g-9.81 m/s2 ing m2 (a) Number (b) Number (c) Number (d) Number Units Units Units Units version 4.24.2.4 es.httr 2000-2017.Jehn.yirasens.Aos. Ar RIghts Reserved. A Division of 2ebnyilevasons, Inc.

Explanation / Answer

Given,

m1 = 478 g ; m2 = 571 g ; R = 4.62 cm ; d = 76 cm ; t = 5.06 s

a)from eqn of motion we know that

s = ut + 1/2 at^2

d = 1/2 at^2 => a = 2d/t^2

a = 2 x 0.76/5.06^2 = 0.06 m/s^2

Hence, a = 0.06 m/s^2

b)for m2

m2g - T2 = m2a

T2 = m2(g - a)

T2 = 0.571 (9.81 - 0.06) = 5.57 N

Hence, T2 = 5.57 N

c)for m1:

T1 - m1g = m1a

T1 = m1 (g + a) = 0.478 (9.81 + 0.06) = 4.72 N

Hence, T1 = 4.72 N

d)we know that, a = alpha R

alpha = a/R = 0.06/0.0462 = 1.29 rad/s^2

Hence, alpha = 1.29 rad/s^2

e)the rotational inertia of the given pulley will be:

I = (T2 - T1)R/alpha

I = (5.57 - 4.72) x 0.0462/1.29 = 0.03044 kg-m^2

Hence, I = 0.03044 kg-m^2

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