A uniform flat disk of radius R and mass M rotates about its center O, on a rough surface with coefficient of friction μ. It's initial angular speed is w0.
Find time duration t before it stops.
M=4kg. R=4m. w0= 10 rad/s.
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Calculate torque dT on a small ring part of radius r and width dr on the disk. Its mass is dm. Integrate to get total torque T.
Let the mass density = k kg/m^2.
T = I *alpha.
I = moment of inertia = M R^2 /2.
M = pi R^2 k
alpha = angular acceleration.
dm = 2 pi r k dr
f = friction force = μ dm g
dT = r × f = (μ dm g)*r = μ 2pi r k dr g* r.
T = 2 pi g μ k R^3 /3
So alpha = 4 g μ/(3 R) = 1/s^2
time t = w0/alpha
= 10 / 1 = 10 sec.
Let the mass density = k kg/m^2.
T = I *alpha.
I = moment of inertia = M R^2 /2.
M = pi R^2 k
alpha = angular acceleration.
dm = 2 pi r k dr
f = friction force = μ dm g
dT = r × f = (μ dm g)*r = μ 2pi r k dr g* r.
T = 2 pi g μ k R^3 /3
So alpha = 4 g μ/(3 R) = 1/s^2
time t = w0/alpha
= 10 / 1 = 10 sec.
kvnmurty:
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Very nice answer , rotation , this is so good chapter in phy . I liked it . thanks for answering .
Thanks@ abhi. Nice of you to appreciate it
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