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 = 0.600 kg, and its center is initially at xy coordinates (

Explanation / Answer

a) We take our positive directions for m1 and m2 to the right and downward respectively. Newton’s 2nd law yields 1: m1a = T, 2:m2a = m2g – T ? a = m2g/(m1 + m2) a = 0.4*9.8(0.6+0.4) a = 3.92 m/s^2 In the coordinate system given in the figure, the accelerations of m1 and m2 are a1 = +3.92i, a2 = -3.92j The acceleration of the center of mass is acom = (m1a1 + m2a2)/(m1 + m2) = 0.6*3.92i-0.4*3.92j/(0.6+0.4) = 2.352 (m/s^2) i-1.568 (m/s^2) j --------> Answer b) The velocity of the center of mass is vcom =vcom,0 + ?acomdt = 2.352ti-1.568tj The center of mass of the system is initially located at xcom,0 = (m1x1,0 + m2x2,0 )/(m1 + m2) = (0.6*-0.5i+0.4*-0.1j)/(0.6+0.4) =-0.3 i - 0.04 j The center of mass of the system is xcom = xcom,0 + ?vcomdt = = (-0.3 +1.176t^2)m i + (-0.04 - 0.784 t^2)m j --------> Answer

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