Asked by Mary
A solid cylinder rolls down an incline plane without slipping. If the center of mass of the cylinder has a linear acceleration of 2.67 m/s2, what is the angle of the incline to the horizontal?
Answers
Answered by
Damon
T = our unknown angle theta
angular acceleration = alpha = d omega/dt
I = (1/2) M R^2
FF = friction force up slope
FF * R = torque
so
FF * R = (1/2) M R^2 * alpha
FF = (1/2)M R alpha
R omega = no slip velocity
R alpha = no slip acceleration = a
so FF = (1/2) M a
F = m a
F = M g sin T - FF = M (2.67)
M g sin T - (1/2)M (2.67) = M(2.67)
9.81 sin T = 1.5 (2.67)
angular acceleration = alpha = d omega/dt
I = (1/2) M R^2
FF = friction force up slope
FF * R = torque
so
FF * R = (1/2) M R^2 * alpha
FF = (1/2)M R alpha
R omega = no slip velocity
R alpha = no slip acceleration = a
so FF = (1/2) M a
F = m a
F = M g sin T - FF = M (2.67)
M g sin T - (1/2)M (2.67) = M(2.67)
9.81 sin T = 1.5 (2.67)
Answered by
Mary
Thanks
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