A 4.00kg rock has a horizontal velocity of magnitude 12.0 m/swhen it is at point

Question

A 4.00kg rock has a horizontal velocity of magnitude 12.0 m/swhen it is at point P in the figure (Figure 1) .

At this instant, what is the magnitude of its angular momentum relative to point O?

What is the direction of the angular momentum in part (A)?

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?

A 4.00kg rock has a horizontal velocity of magnitude 12.0 m/s when it is at point P in the figure (Figure 1) . At this instant, what is the magnitude of its angular momentum relative to point O? What is the direction of the angular momentum in part (A)? 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? What is the direction of the rate in part (C)?

Explanation / Answer

Mass of rock = 4 kg

velocity of rock = 12 m/s

radius = 8 m

v in tangential direction = v * sin (36.9)

Angular Momentum = m * v * sin(36.9) * r

Angular Momentum = 4 * 12 * sin(36.9) * 8

a) Angular Momentum = 230.56 kgm^2/s

Direction of the angular momentum can be given by right hand rule

b) So the direction of the angular momentum is into the page

Torque = F * r * cos (36.9)

Torque = m * g * r * cos (36.9)

Torque = 4 * 9.8 * 8 * cos (36.9)

Torque = 250.78 Nm

Rate of change of Angular Momentum = Torque

c) Rate of change of Angular Momentum = 250.78 Nm

Direction of the angular momentum can be given by right hand rule

d) So the direction of the angular momentum is out of the page

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