1. A thin ring of radius R= 1 cm rotates about a vertical axis passing through i
ID: 2258731 • Letter: 1
Question
1.
A thin ring of radius R=1 cm rotates about a vertical axis passing through its
center with frequency f = 20 Hz. It is then placed flat on a table. The coefficient of
friction between the ring and the table is ?=0.1. a) After what time will the ring
stop spinning? b) How many revolutions will it make?
2.
Forces F1 = 2 N, F2 = 2.5 N and F3 = 1 N are
applied to a rod of length L=1 m and mass m=1 kg
. There are no other forces
(also, no gravity). Through which angle will the
rod rotate after 1 second?
F1 applied to the right, on top of the rod, F2 applied 25cm below F1 to the left, F3 applied at the bottom of the rod to the right
Explanation / Answer
First assume that ring rotates about a fixed axis about it scenter.
Now every point on the rim will have velocity ?r. and it willget to zero after time, say, t.
so we can use the first equation of rotationalkinematics, wf = wi + alpha*t
now we only need to find alpha, which can be found out asfollows
lets say its mass is "m". then the friction force acting on thewhole rim will be "mu*mg" in direction opposite to velocity ,tangentially. which implies that the acceleration will be"-mu*g" (= F/m) (as force is opposite to direction of velocity,assumed +ve direction)
Now, alpha = acceleration / radius = -mu*g / r
also wf = 0
=> 0 = wi + alpha*t
=> w + (-mug/r)t = 0
=> t = wr/mu*g
w=2*pi*f=2*3.142*20=125.68 rad/s
t=wr/mu*g=1.282 sec
after this much time the ring will stop.
for revolutions, we need to find theta.
theta = wi * t + (alpha*t*t/2)
on substituting,
theta = w^2r/2mu*g
=> revolutions = theat/2?
=> revolution = w^2r/4?*mu*g
no.of revolutions=12.82
problem 2
Torque due to all forces..Tnet=T1+T2+T3...
T1=-2*0.5=-1N/m since it is moves in clockwise
T2=2.5*0.25=0.625 it moves in anti clockwise
T3=1*0.5=0.5N/m since it is moves in anti clockwise
Tnet=0.125 N.m in anti clockwise direction
Tnet=I*alpha...
I is moment of inertia...=(1/12)*M*L^2=0.0834 kg.m^2...
alpha=Tnet/I=1.5 rad/s^2....
dispalcement theta=(1/2)*alpha*t^2=0.749 rad
theta=0.749*360/(2*pi)=42.931 degrees from the equilibrium
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