1. Two uniform metal disks, one with radius R1 = 2.50cm and mass M1 = 0.80 kg an
ID: 1379582 • Letter: 1
Question
1. Two uniform metal disks, one with radius R1 = 2.50cm and mass M1 = 0.80 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 kg, are welded together and mounted on a frictionless axle through their common center. A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is Suspended from the free end of the string. If the block is released From rest at a distance of 2.00 m above the floor, what is its Speed just before it strikes the floor? (Note: The moment of Inertia of a uniform solid disk is given by (1/2)MR^2).Explanation / Answer
here
m1 = 0.80 kg
m2 = 1.60 kg
r1 = 0.025 m
r2 = 0.05 m
M = 1.5 kg
d = 2 m
total inertia
Inet = 1/2 * (m1 * r1^2 + m2 * r2^2)
then by using
Newton's 2nd law for the hanging block
M*g - T = M*a ......................(1)
The only force causing a torque is T.
T* r1 = Inet *alpha
No-slip condition
alpha = a / r1
T * r1^2 = Inet * a
T = Inet*a/r1^2.........................(2)
then put the equation (2) in the equation (1)
M*g - Inet*a/r1^2 = M*a
M*g = a*(M + Inet/r1^2)
a = g*M*r1^2/(M*r1^2 + Inet)
then by using third equation of motion
vf^2 = vi^2 + 2*a*d
vi = 0, thus:
vf^2 = 2*a*d
So,
vf = sqrt(2*a*d)
by puttiing the value of a
vf = sqrt(2*d*g*M*r1^2/(M*r1^2 + Inet))
then by putting the equation of Inet
vf = sqrt(2*d*g*M*r1^2/(1/2*(m1*r1^2 + m2*r2^2) + M*r1^2 ) )
vf = 2*sqrt(d*g*M*r1^2/(m1*r1^2 + m2*r2^2 + 2*M*r1^2) ).................(3)
then put all the values
vf = 2 * sqrt( 2 * 9.8 * 1.5 * 0.025^2 / 0.8 * 0.025^2 + 1.6 * 0.05^2 + 2 * 1.5 * 0.025^2))
vf = 3.395 m/s
the speed just before it strikes the floor is 3.395 m/s
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