Rotational Dynamics and Static Equilibrium. What is the acceleration of the ccart?
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A 350 g air track cart on a horizontal air track is attached to a string that goes over a pulley with a moment of Inertia of 6 x 10^-6 kg.m^2 and the radius of 1.35 cm. The string is pulled vertically downwards by a force of 2.5 N. What is the acceleration of the cart? (a) 7.14 m/s^2 (b) 6.52 m/s^2 (c) 5.27 m/s^2 (d) 4.98 m/s^2 I believe the answer is B. How do you work out this problem? thNks
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Answer:
force = m*a + I*alpha the first part is for the cart and the second is for the pulley Then relate the angular acceleration to the linear acceleration as follows: alpha=angular acceleration= w^2*r in rad/s so a(accleration of cart)= w^2*r^2 m/s so force=m*a + I*(a/r) from which you can find a (acceleration)
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Other answers
For the above system, torque=F*R about the axis of the pulley. but torque = torque on pulley+torque on cart = Ia/R + maR = FR i.e. a=FR/(I/R + mR) plugging in the values, a=6.52 m/s2
astronikil
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