Question

A 7.1 kg, 5.0 m board is balanced about a point 2.0 m from right end of the board. A cube is sitting centered 0.5 m from the right end. A 3.5 kg cylinder is dropped from a height of 2.0 m onto the left end of the board and sticks to it. The dropped cylinder is going 6.26 m/s just before it hits the left end of the board. (Assume the cube and cylinder are point masses.)

a) What is the mass of the cube sitting 0.5 m from the right end?

b) What is the cylinder’s angular momentum about the board/cube's pivot point just before hitting the board?

c) What is the total angular momentum of all of the masses just after the collision?

d) What is the moment of inertia of the board and two masses just after collision?

e) What is the board's angular velocity just after collision?

f) How fast is the cube (originally 0.5 m from the right end) going just after the collision?

if you need a diagram, draw one!!

Answer 1

Part (A) By balancing torque about point 'p' (p_int) m_1g (2 - 0.5) = Mg (2.5 - 2) m_1 times 1.5 = 7.1 times 0.5 m_1 = 2.36 kg Part (B) Angular momentum of cylinder L_1 = IW = m_2R^2 w= mvR L_i = 3.5 times (2.5 times 0.5) times 6.26 L_i = 65.73 kg- m^2/ s^2 Part (C) Total circular momentum after collision remain same as angular momentum before collision s', L_f = 65.73 kg-m^2/s Part (d) Moment of Inertia I_bboard = (ML^2/12 + Md^2) I_board = 7.1 times (5)^2/12 + 7.1 (0.5)^2 implies I_board = 16.54 kg-m^2 I_masses= m_1d_1^2 + m_2 d_2^2 = 2.36 times (1.5)^2 + 3.5 times (3)^2 I_masses = 5.31 + 31.5 implies I_mass = 36.8 kg- m^2 Part (e) Just after collision L_i = L_f 65.73 = ((I_board) + I_mass) W w = 1.23 rad/ s Part (f) for cube v = wd,