Rotational Dynamics and Static Equilibrium. What is the acceleration of the ccart?

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

Rotational Dynamics and Static Equilibrium. What is the acceleration of the ccart?
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)
Reply: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

anemone