The figure below shows an arrangement with air track. In which a cart is connect
ID: 1571888 • Letter: T
Question
The figure below shows an arrangement with air track. In which a cart is connected by a cord to a hanging block. The cart has mass m_1 = 0.520 kg and its center is initially at xy coordinates (-0.500 m, 0 m). The block has mass m_2 = 0.240 kg and its center is initially at xy coordinates (0, -0.100 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. In unit-vector notation what is the acceleration of the center of mass of the cart-block system? (Express your answer in vector form.) m/s^2 What is the velocity of the center of mass as a function of time t? (Assume t is in seconds.) (Express your answer in vector form.) vector v_com = m/s Sketch the path taken by the system's center of mass. (Submit a file with a maximum size of 1 MB.) This answer has not been graded yet. Which of the following describes the path taken by the system's center of mass?Explanation / Answer
a)Using newtons second law
The force action on m1 in x-direction
T =m1a
the force acting on m2 in ydirection
m2a=m2g-T
m2a=m2g-m1a
a(m1+m2) =m2g
a=m2g/(m1+m2)
a =0.40*9.8/(0.40+0.66)
a=3.69 m/s^2
in coordinate system the accelerations of m1 and m2 are
a1=3.69i
a2=-3.69j
the acceleration of center of mass is
a_com =(m1a1+m2a2)/(m1+m2)
a_com =(0.66*3.69-0.40*3.69)/(0.40+0.66)
a_com =2.29i-1.39j
b)
the velocity of center of mass
V_com=2.29ti-1.39tj
the ceter of mass of system initially located at
X_comi =(m1x1i+m2y2i)/(m1+m2)
X_comi =-0.23 i -0.11 j
the center of mass of system is
X_com =X_comi +integral[V_com dt]
X_com =(-0.23+1.145t^2) i + (-0.11-0.695t^2) j
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