Question
A ball of mass 0.1 kg is suspended vertically from the ceiling with a "massless" string and
brought to rest. The string is one meter long. When a force of 0.5N is applied 45 deg. below
the horizontal, what will be the magnitude of the resulting angular acceleration of the ball?
How would you go about solving this?
brought to rest. The string is one meter long. When a force of 0.5N is applied 45 deg. below
the horizontal, what will be the magnitude of the resulting angular acceleration of the ball?
How would you go about solving this?
Answers
The torque is M=F•cosα•L.
The ball is material point => its moment of inertia is
I=m•L².
The Newton’s 2 law for rotation
M=I•ε,
F•cosα•L= m•L²•ε,
ε= F•cosα•L/ m•L²=F•cosα /m•L.
The ball is material point => its moment of inertia is
I=m•L².
The Newton’s 2 law for rotation
M=I•ε,
F•cosα•L= m•L²•ε,
ε= F•cosα•L/ m•L²=F•cosα /m•L.
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