A spool of thread consists of a cylinder of radius R1 = 6.8 cm with end caps of
ID: 1286784 • Letter: A
Question
A spool of thread consists of a cylinder of radius R1 = 6.8 cm with end caps of radius R2 = 11.5 cm as depicted in the end view shown in the figure below. The mass of the spool, including the thread, is m = 240 g. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force T = 0.700 N acting to the right is applied to the free end of the thread. For the moment of inertia treat the spool as being a solid cylinder of radius R1, as the extended edges are thin and therefore light.
What is the acceleration of the spool?
Explanation / Answer
Clearly the spool's acceleration will be in a horizontal direction (it won't jump vertically off the table); so that means the net force must be horizontal. And that means we can ignore any vertical forces (gravity, normal force) for now.
The only two horizontal forces are:
* The tension: +T
* Friction from the table: ?Ff (negative because it acts toward the left)
so:
a = Fnet / m = (T?Ff)/m
But the problem is, we don't know the value of "Ff". So we need some more equations, and since the spool rotates, it's probably a good idea to use equations of torque.
The two horizontal forces produce torques around the spools axis:
* Ff produces a clockwise torque: (Ff)(R2)
* T produces a counter-clockwise torque: ?T(R1)
So the net torque is:
?_net = (Ff)(R2)?T(R1)
Now use: ?_net = I? (this is the rotational equivalent of: Fnet = ma)
(Ff)(R2)?T(R1) = I?
But if it's rolling without slipping, then ? = a/R2. That means:
(Ff)(R2)?T(R1) = Ia/R2
Recap:
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You now have these two equations:
a = (T?Ff)/m
(Ff)(R2)?T(R1) = Ia/R2
These are two equations in two unknowns ("a" and "Ff"). Use the algebra of simultaneous equations to eliminate "Ff" and solve for "a" in terms of "T", "m", "R1" and "R2".
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