Gravitational acceleration = g = 9.81 m/s2
Mass of the disk = M = 6 kg
Radius of the disk = R = 0.25 m
Moment of inertia of the disk = I
I = MR2/2
I = (6)(0.25)2/2
I = 0.1875 kg.m2
Mass of the crate = m = 18 kg
Angle of the incline = = 30o
Normal force on the crate from the incline = N
Coefficient of kinetic friction between the crate and the incline = k = 0.24
Friction force on the crate = f = kN
Tension in the rope = T
Angular acceleration of the crate =
Acceleration of the block = a
a = R
From the free body diagram of the crate,
N = mgCos
f = kmgCos
ma = mgSin - f - T
ma = mgSin - kmgCos - T
T = mgSin - kmgCos - ma
For the disk,
I = TR
I = (mgSin - kmgCos - ma)R
I = mgRSin - kmgRCos - m(R)R
(I + mR2) = mgR(Sin - kCos)
[0.1875 + (18)(0.25)2] = (18)(9.81)(0.25)[Sin(30) - (0.24)Cos(30)]
= 9.8 rad/s2
Magnitude of angular acceleration of the disk = 9.8 rad/s2
1. (P) The figure below shows an 18.0kg crate of salami sliding down a ramp on...
1. (P) The figure below shows an 18.0kg crate of salami sliding down a ramp on a (massless) rope that is wrapped around a disk of radius 0.25m and mass 6.0kg. The coefficient of kinetic friction between the crate and ramp is μ k-0.24. What is the magnitude of the disk's angular acceleration? 30° A 3.1 rad/s2 B 6.7 rad/s2 C 9.8 rad/s2 D 11.4 rad/s2 E 13.7 rad/s2
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