In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding
ID: 1552209 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding block has a mass of m2 = 0.840 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
m2 is on a flat table
(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away.
(b) Find the angular speed of the pulley at the same moment.
Explanation / Answer
m1 = 0.495 kg
m2= 0.840 kg
M= mass of pulley = 0.350 kg
inner radius R1 = 0.0200 m
R2 = 0.0300 m
uk= cofficient of kinetic friction = 0.250
Vi = initial velocity = 0.820 m/s
a) Vf( speef after it moved d= 0.700m) = ?
from work energy theorem:
change in K.E ( m1 + m2 + pulley) = net work done
(1/2)*( m1+m2) ( Vf2 - Vi2) + (1/2)*I ( wf2 - wi2) = ( m1*g - uk*m2*g ) d ---------------------1
[ here f( friction force on m2) = uk*m2*g
I= moment of interia = (1/2)*M*( R12 + R22 )
w = V/ R2]
putting this in eq 1:
(1/2)*( m1+m2) ( Vf2 - Vi2) + (1/2) *( 1/2)*M*( R12 +R22) ([ Vf2 - Vi2] / R22) = ( m1*g - uk*m2*g)*d
on rearranging the terms:
Vf = ( Vi2 + [ ( m1g- uk*m2*g)*d / (1/2)*(m1+m2) + (1/2)*M *( 1+ R12 / R22 )] )1/2
Vf = ( 0.8202 + [ 0.495*9.8 - 0.250*0.840*9.8 ) *0.700 / (1/2)*( 0.495+ 0.840) + (1/2)*0.350*( 1+ 0.02002 / 0.03002) ])1/2
Vf = 1.67 m/s
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b) wf ( angular speed) = Vf/R2
wf = 1.67 / 0.0300 =55.67 rad/s