A 0.431 kg ball is located on the x-axis 3.77 m from the origin. It\'s launched

Question

A 0.431 kg ball is located on the x-axis 3.77 m from the origin. It's launched straight up with a speed of 52.9 m/s.
a) What is its angular momentum about the origin at maximum height?
b) What is its angular momentum about the origin halfway back down?
c) What is its torque due to gravitational force about the origin at maximum height?
d) What is its torque due to gravitational force about the origin halfway back down?

A 0.431 kg ball is located on the x-axis 3.77 m from the origin. It's launched straight up with a speed of 52.9 m/s.
a) What is its angular momentum about the origin at maximum height?
b) What is its angular momentum about the origin halfway back down?
c) What is its torque due to gravitational force about the origin at maximum height?
d) What is its torque due to gravitational force about the origin halfway back down?


a) What is its angular momentum about the origin at maximum height?
b) What is its angular momentum about the origin halfway back down?
c) What is its torque due to gravitational force about the origin at maximum height?
d) What is its torque due to gravitational force about the origin halfway back down?

Explanation / Answer

a)

At maximum height it velocity = 0

So, its angular momentum = 0

b)

maximum height reached,

h = u^2/2g = 52.9^2/(2*9.8)

= 142.8 m

So, height needed, h' = h/2 = 71.4 m

So, for the velocity at h', using the equation of motion:

v^2 = u^2 + 2as

So, v^2 = 52.9^2 - 2*9.8*71.4

So, v = 37.4 m/s

So, angular momentum = m*37.4*3.77

=  0.431*37.4*3.77

= 60.8 kg.m^2/s

c)

Torque due to gravitational force at max height,

T = mg*3.77

= 0.431*9.8*3.77

= 15.9 N.m

d)

At half way back down, torque

= 0.431*9.8*3.77

= 15.9 N.m

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