The figure shows an arrangement with an air track, in which a cart is connected

Question

The figure 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 = 0.660 kg, and its center is initially at xy coordinates (–0.380 m, 0 m); the block has mass m2 = 0.400 kg, and its center is initially at xy coordinates (0, –0.300 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?

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.