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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Sep 4th 2008, 06:53 AM #1 Junior Member   Join Date: Sep 2008 Posts: 7 Rotational Motion A constant horizontal force F of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 10 kg, its radius is 0.1 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (The answer is 1.6m/s^2) (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (The answer is 16rad/s^2) (c) In unit-vector notation, what is the frictional force acting on the cylinder? (The answer is (4.0N)i ) I tried with τ (Torque) = I*a (alpha) (12 - f)(r) = Ia/r 12- f = 0.05a/(0.1^2) I have 2 unknowns. What shall I do? The answers that I typed are the correct ones, I would like to know that solution. Thank you very much.   Sep 4th 2008, 07:41 AM   #2

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 Originally Posted by noppawit A constant horizontal force F of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 10 kg, its radius is 0.1 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (The answer is 1.6m/s^2) (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (The answer is 16rad/s^2) (c) In unit-vector notation, what is the frictional force acting on the cylinder? (The answer is (4.0N)i ) I tried with τ (Torque) = I*a (alpha) (12 - f)(r) = Ia/r 12- f = 0.05a/(0.1^2) I have 2 unknowns. What shall I do? The answers that I typed are the correct ones, I would like to know that solution. Thank you very much.
Is the friction force really providing a torque here? Recall that the axis of rotation is not the center of the wheel, but where the wheel makes contact with the surface. So
$\displaystyle \sum \tau = (12~N)(2 \cdot 0.1~m) = \left ( \frac{1}{2} \cdot (10~kg) \cdot (0.1~m)^2 \right ) \left ( \frac{a}{0.1~m} \right )$

-Dan
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