You see some management students getting drunk at 1 PM in the quad. (Yes, this is an early wake up
call for them.) You decide to tie a 2.00 daN water balloon on the end of a 1.00 m long light rope and
start swinging it horizontally. The rope from your hand to the water balloon makes an angle of 30.0
o
below the horizontal.
(a) What is the tension in the rope?
(b) What is the speed of the water balloon?
(c) Suppose you double the speed of the water balloon given in part (b) over four seconds. Assume
the acceleration is uniform. Just as you are approaching the four second mark, what is the net acceleration of the water balloon? What angle is the rope making with respect to the horizontal now?
(d) Suppose you release the rope and the water balloon flies off at the speed given in part (c) and at
an angle of 45.0 above the horizontal. What is the range of the water balloon if it left from a height of 1.75 m above the ground?
What a bad question. What do the drunken students have to do with any of this? And what's a daN?
Sigh...
Well it says horizontally so I guess that means over one's head. If so:
T sinθ = mg and I assume the daN thing is mg.
b)T cosθ = mv^2/r
You'll need to find r using a little trig, then solve for v
c) Now things get even murkier. Net? does he mean angular (change in omega over t)? Centripetal (v^2/r)? As for the angle do part b in reverse with the new (tangential I guess) velocity.
d) how did it suddenly get to 45o? Do we just ignore everything else and treat it as a straight projectile motion question? Who knows....
To answer these questions, we need to apply the principles of Newton's laws of motion, circular motion, and projectile motion. Let's break down each part step by step:
(a) To find the tension in the rope, we need to analyze the forces acting on the water balloon at the bottom of its swing. The tension in the rope provides the centripetal force required to keep the balloon moving in a circle. The gravitational force acts vertically downward. We can start by resolving these forces.
Since the rope makes an angle of 30.0 degrees below the horizontal, the vertical component of tension balances the gravitational force, and the horizontal component of tension provides the centripetal force.
The tension in the rope can be calculated as follows:
Tension (T) = Gravitational force (mg) - Vertical component of tension
(b) Once we have the tension in the rope, we can calculate the speed of the water balloon using the centripetal acceleration formula. The centripetal acceleration is given by:
Centripetal acceleration (ac) = (v^2) / r
Here, r is the length of the rope, and v is the speed of the water balloon. We can rearrange the equation to solve for v:
v = sqrt(ac * r)
(c) If we double the speed of the water balloon over four seconds with a uniform acceleration, we need to find the acceleration first. We can use the equation for uniformly accelerated motion:
Final velocity (v) = Initial velocity (u) + (acceleration * time)
Rearranging the equation to solve for acceleration:
Acceleration = (v - u) / time
After calculating the acceleration, we can find the net acceleration of the water balloon just before the four-second mark.
To find the angle the rope makes with respect to the horizontal, we can calculate the arc tangent of the vertical component of tension divided by the horizontal component of tension.
(d) Lastly, to find the range of the water balloon, we can use the range formula for projectile motion:
Range (R) = (v^2 * sin(2θ)) / g
Here, v is the speed of the water balloon, θ is the angle of projection, and g is the acceleration due to gravity. We can substitute the given values to find the range.
Note: Some values and measurements may need to be converted to SI units before performing the calculations.
Now, armed with this explanation, you can use the appropriate equations and plug in the given values to find the answers to each part of the problem.