FULL SCREEN PRINTER VERSION BACK NEXT Question1 In the figure, a 0.400 kg ball i

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FULL SCREEN PRINTER VERSION BACK NEXT Question1 In the figure, a 0.400 kg ball is shot directly upward at initial speed 60.7 m/s. What is its angular momentum about P, 6.06 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground? Ball (a) Number Units (b) Numberunits (c) Number Units (d) Number Units Question Attempts: 0 of 12 used SAVE FOR LATER SUBMIT ANSWER

Explanation / Answer

We know that angular momentum is

L=m(r*V)

where r*V is the cross product between r and V vector.

So,

a)

When the ball is at the highest point v=0, so

Angular momentum will also be zero.

b)

For this part we have to find the max height,

02-60.72=2*(-9.81)H

H=187.79 m

So half way the max will be H/2

=93.896m

Now velocity at this point will be,

v2=2g(H/2)

v=42.921m/s

So angular momentum will be

L=0.4(6.06*42.921)

=104.04143 kg-m2/s

c)

Torque is

r*F

So,

gravtional force is same at all the points and r is also same.

So

Torque=6.06*(.4*9.81)

=23.77944 N-m

d)

It iwll be same as C, as told earlier.

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