Question

a 350 G AirTrack cart on a horizontal Air Track is attached to a spring that goes over a pulley with a moment of inertia of 6 x 10 to the negative ^6 kilogram * meter squared and a radius of 1.35 CM the string is pulled vertically downward by a force of 2.5 Newtons what is the acceleration of the cart?

Answer 1

Let T be the tension in the string between the pulley and the cart.

Force balance on the cart:

T - m*a = 0 (a is the acceleration of the cart)

Force balance on the pulley (F - T)*r - α*I = 0 (r = pulley radius, F the applied force, α the angular acceleration of the pulley and I its moment of inertia).

However, α = a/r where a = tangential acceleration of the pulley, which must be the same as the acceleration of the cart and string. The second equation becomes

(F - T)*r - (a/r)*I = 0

You know have 2 eq in 2 unknowns (a and T)

T - m*a = 0
(F - T)*r - (a/r)*I = 0

solve for a from the second

a = (F - T)*r²/I

and put it in the first

T - m*(F - T)*r²/I = 0

solve for T

T*(1 + m*r²/I) = m*F

T = (m*r²/I)*F/(1 + m*r²/I)

m*r²/I = .350*.0135²/0.000006 = 10.63

T = 10.63*2.5/11.63 = 2.29 N

a=(2.5-2.29)*0.0135*0.0135/6*10^-6

a=6.38 m/s2