Asked by John
A small disk of radius R1 is mounted coaxially with a larger disk of radius R2. The disks are securely fastened to each other and the combination is free to rotate on a fixed axle that is perpendicular to a horizontal frictionless table top. The rotational inertia of the combination is I. A string is wrapped around the larger disk and attached to a block of mass m, on the table. Another string is wrapped around the smaller disk and is pulled with a force F as shown. The acceleration of the block is:
Answers
Answered by
Delaram
FR1-TR2=I*alfa / alfa=rotational acceleration)
alfa=a/R2
so we can find T
T=FR1/R2-Ia/(R2^2)
then we use T=ma to find a:
a=FR1R2/(mR2^2+I)
alfa=a/R2
so we can find T
T=FR1/R2-Ia/(R2^2)
then we use T=ma to find a:
a=FR1R2/(mR2^2+I)
Answered by
Delaram
in other words we must use second law of newton to find acceleration, a=T/m
but since we don't have tension we should utilise torque equation to find T, the point here is that we should notice angular acceleration=a/R2
but since we don't have tension we should utilise torque equation to find T, the point here is that we should notice angular acceleration=a/R2
Answered by
Anonymous
0.15
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