Cannons and projectiles
A 30 kg object, A, was fired from a cannon in projectile motion. When the
projectile
was at its maximum height of 25 m, its speed was 20 m/s.
An identical object, B, was attached to a mechanical arm and moved at a constant
speed of 20 m/s in a vertical half-circle. The length of the arm was 25 m.
Ignore air
resistance.
(a) Calculate the force acting on object A at its maximum height.
(b) Calculate the time it
would take object A to reach the ground from its position
of maximum height.
(c) Describe and compare the vertical forces acting on objects A and
B at their
maximum heights.
Answer:
(a) Force on projectile = mg = 30x9.81 =
294.3 N
(b) Maximum height = 25 m using s = ut + ½ at
2 and with the initial
vertical speed at the top u = 0 we have:
25 = ½ gt
2 therefore t =
[50/9.8]
1/2 = 2.25 s
(c) The object on the mechanical arm is subject to a
centripetal force to keep it moving in a circle. This force (F) is given by the equation:
F =
mv
2/r = [30x400]/25 = 480 N which is greater than the weight of the object. Some of
the force will be provided by the weight of the object.
The additional force of the arm on
the object needed to keep it moving in a circle is therefore 480 - 294.3 = 185.7
N
I have worked out the angle of projection as well (you don't actually need
that!)
At maximum height its vertical speed is zero (if it wasn't it wouldn't be the maximum
height).
The horizontal speed is constant and equal to 20 m/s.
Take the initial
velocity to be v at an angle to the horizontal of A
and so 20 = v cos A 400 =
v
2cos
2A
Vertically: initial velocity = vsin A and 0 =
v
2sin
2A – 2x9.81x25
Combining the horizontal and vertical
equations:
400tan
2A = 2x9.8x25 this gives A = 47.9
o
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