A student has built the catapult shown below in an attempt to launch his physics
ID: 1454587 • Letter: A
Question
A student has built the catapult shown below in an attempt to launch his physics book into the middle of a lake. Initially he secures the launch arm to the ground using a rope as shown on the left. Then he releases the rope and the catapult arm swings and launches the book as shown on the right. The catapult arm is made from a massless board, and the pivot point is placed with L1 = 1 m and L2 = 4 m. The counterweight has mass m1 = 100 kg and the physics book has mass m2 = 1 kg.
a. What is the tension T on the rope before he releases the arm (i.e. with the rope still attached)?
b. What is the magnitude of the force exerted by the pivot on the board before he releases the arm (i.e. with the rope still attached)?
c. What is the direction of the total torque after the rope is detached?
d. What is the moment of inertia of the arm rotating on its pivot with the masses attached? Treat m1 and m2 as point particles, and remember that the board is massless.
e. What is the magnitude of the angular acceleration right after he releases the arm? Answer Range: [5.0 10.0 rad/s2 ]
Please show a step by step solution for this, I have not worked with a problem like this before, thank you!
m2 2 m2 mi RopeExplanation / Answer
a)
In equilibrium net torque about the pivot = 0
m2*g*L2 + T*L2 = m1*g*L1
(1*9.8*4) + (T*4) = 100*9.8*1
T = 235.2 N <<------------answer
(b)
along vertical
Fnet = 0
F - T - m1*g - m2*g = 0
F = (235.2 + (110*9.8))
F = 1313.2 N
(c)
clock wise
(d)
I = m1*L1^2 + m2*L2^2 = 100*1^2 + 1*4^2 = 100+16 = 116 kg m^2
(e)
net torque = I*alpha
(m1*g*L1 - m2*g*L2 = I*alpha
(100*9.8*1) - (1*9.84) = 116*alpha
angular acceleration = alpha = 8.4 rad/s^2
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