A 0.400 kg ball is shot directly upward at initial speed 40.0 m/s. What is its angular momentum about a point 2.0 m horizontal 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 the point 2m from the horizontal launch point due to the gravitational force when the ball is c) at maximum height and d) halfway back to the ground?
I know that because max height is v=0, there is 0 angular momentum, but i can't figure out the rest
Torque is always (2m)xF=-2mg
velocity at h/2 is -vo/root2=-28.284
So Angular momentum is
(2m)xmv=-22.627
THANK YOU SO MUCH!
asd
To find the angular momentum of the ball about a point 2.0 m horizontally from the launch point, we first need to calculate the ball's linear momentum and then determine its rotational motion.
a) At maximum height:
- At the maximum height, the vertical component of the ball's velocity is zero, but the horizontal component remains constant.
- Since the ball is not moving horizontally at this point, its linear momentum with respect to the point 2.0 m from the launch point is zero.
- Angular momentum (L) is given by L = r x p, where r is the position vector from the point to the object, and p is the linear momentum vector.
- As the linear momentum is zero, the angular momentum is also zero.
b) Halfway back to the ground:
- At this point, the ball is moving both vertically and horizontally.
- To calculate the angular momentum, we need the linear momentum and the position vector.
- The linear momentum (p) is given by p = m * v, where m is the mass of the ball and v is its velocity.
- Therefore, p = 0.400 kg * 40.0 m/s = 16.0 kg·m/s.
- The position vector (r) is a vector that points from the point 2.0 m horizontally from the launch point to the ball's position at this halfway point.
- Since the ball is moving vertically, the position vector is a straight vertical line.
- The magnitude of the position vector is 2.0 m.
- The angular momentum (L) is then given by L = r x p = r * p * sin(θ), where θ is the angle between the vectors r and p.
- In this case, the angle θ is 90 degrees because the linear momentum is vertical and the position vector is vertical.
- Therefore, the angular momentum is L = 2.0 m * 16.0 kg·m/s * sin(90°) = 32.0 kg·m²/s.
c) At maximum height, the torque on the ball due to the gravitational force:
- The torque (τ) is given by τ = r × F, where r is the position vector and F is the force vector.
- In this case, the force acting on the ball is the gravitational force, with magnitude F = m * g, where m is the mass of the ball and g is the acceleration due to gravity.
- The position vector (r) points from the point 2.0 m from the launch point to the ball's position at maximum height.
- The magnitude of the position vector is 2.0 m.
- Therefore, the torque is τ = 2.0 m × (0.400 kg × 9.8 m/s²) = 7.84 N·m.
d) At halfway back to the ground, the torque on the ball due to the gravitational force:
- The position vector and the force vector have the same direction (vertical).
- Therefore, the angle between the vectors is zero, and sin(θ) becomes 0 (sin(0°) = 0).
- As a result, the torque at this point is zero.
So, the answers are:
a) Angular momentum = 0 kg·m²/s
b) Angular momentum = 32.0 kg·m²/s
c) Torque = 7.84 N·m
d) Torque = 0 N·m
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Do the second question first. (It is worded incorrectly I think because there is in fact no torque on the ball since the force goes right through its center. There is however a torque about that point 2 meters from the launch point) The torque is the same at the top and halfway down because the perpendicular distance from the force vector to the launch point is always 2 meters. The magnitude of the torque = m g (2)
Well, we better figure out how fast it is going when it is halfway down. You already answered for at the top, zero because velocity is zero.
a = - 9.8
v = 40 - 9.8 t
h = 0 + 40 t - 4.9 t^2
when h = 20
20 = 40 t - 4.9 t^2
4.9 t^2 - 40 t + 20 = 0
t = (1/9.8)[ 40 +/- sqrt(1600 -392)]
t = (1/9.8) [ 40 +/- 34.8 ]
t = 7.63 on the way down (the plus sign)
v = 40 - 9.8(7.63) = -34.8 m/s
angular momentum = m V x R
= m (-34.8)(horizontal distance which is 2)
= -69.6 * mass