Exercise 10.37 A 2.20 kg rock has a horizontal velocity of magnitude 12.0 m/s wh
ID: 1478311 • Letter: E
Question
Exercise 10.37
A 2.20 kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure (Figure 1) .
Part A
At this instant, what is the magnitude of its angular momentum relative to point O?
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Part B
What is the direction of the angular momentum in part (A)?
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Part C
If the only force acting on the rock is its weight, what is the magnitude of the rate of change of its angular momentum at this instant?
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Part D
What is the direction of the rate in part (C)?
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Figure 1 of 1
Exercise 10.37
A 2.20 kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure (Figure 1) .
Part A
At this instant, what is the magnitude of its angular momentum relative to point O?
L = kgm2/sSubmitMy AnswersGive Up
Part B
What is the direction of the angular momentum in part (A)?
What is the direction of the angular momentum in part (A)? into the page out of the pageSubmitMy AnswersGive Up
Part C
If the only force acting on the rock is its weight, what is the magnitude of the rate of change of its angular momentum at this instant?
dL dt = kgm2/s2SubmitMy AnswersGive Up
Part D
What is the direction of the rate in part (C)?
What is the direction of the rate in part (C)? into the page out of the pageSubmitMy AnswersGive Up
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Figure 1 of 1
Explanation / Answer
a)
L = m r X v that is cross product of two vectors r and v
L = mvr sin(angle) = (2.2)(12)(8)sin(143.1degrees) = 126.8 kgm^2/s
b) Direction is into the page
c)
dL/dt = d[mvr sin(angle)] /dt = r d[mv sin(angle)] /dt
= (2.2)(12)(8)cos(143.1degrees) = (-168.89) kgm^2/s
dL/dt = r X F which is torque
d) Since there is negative sign direction is out of the page
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