Figure 9-44 shows an arrangement with an air track, in which a cart is connected

Question

Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass m1 = 0.630 kg, and its center is initially at xy coordinates (–0.470 m, 0 m); the block has mass m2 = 0.540 kg, and its center is initially at xy coordinates (0, –0.250 m). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart–block system? (b) What is the velocity of the com as a function of time t, in unit-vector notation?

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Explanation / Answer

First we need to find the acceleration of the system. To do this, draw two separate force body diagrams on each cart.

For cart 1:
Fx = T = m1(+a)

For cart 2:
Fy = T - Fg2 = m2(-a)

Plug in the tension from cart 1 into the other FBD equation (because the tension is the same throughout the system) and solve for a:
m1a - m2g = m2(-a)
a(m1 + m2) = m2g
a = (m2g)/(m1 + m2)
a = ( .54)(9.8)/(.63 + .54)
a = 4.523076923 m/s2

Therefore the acceleration of mass 1 is 4.523i and the acceleration of mass 2 is -4.523j due to the direction that they are moving. From here we can find the acceleration of the center of mass:
acom = miai / mi
acom = (m1a1 + m2a2) / (m1 + m2)
acom = (.63(4.523i) + (.54(-4.523j)) / (.63 + .54)
acom = (2.849538i - 2.4424615j) / (1.17)
acom = 2.4355i - 2.08757j

In order to find the velocity of the center of mass, we integrate the acceleration of the center of mass:
vcom = acomdt
vcom = 2.4355ti - 2.0876t j

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