In the figure, a 125 g ball on a 75.1 cm rope is swung in a vertical circle about a point 267 cm above the ground. If the ball is swung at the slowest speed where the ball goes over the top of the circle without the rope having any slack and the rope is released when the ball is at the top of the loop, how far to the right does the ball hit the floor?
Question
In the figure, a 125 g ball on a 75.1 cm rope is swung in a vertical circle about a point 267 cm above the ground. If the ball is swung at the slowest speed where the ball goes over the top of the circle without the rope having any slack and the rope is released when the ball is at the top of the loop, how far to the right does the ball hit the floor?
Solution
To solve this problem, we need to use the principles of physics, specifically the conservation of energy and the equations of motion.
Step 1: Determine the height of the ball when it is at the top of the circle. This is the sum of the length of the rope and the height of the point above the ground, which is 75.1 cm + 267 cm = 342.1 cm or 3.421 m.
Step 2: Calculate the speed of the ball at the top of the circle. Since the rope is just taut at the top of the circle, the ball's speed is just enough to provide the centripetal force required for circular motion. This means that all of the ball's gravitational potential energy (mgh) at the top of the circle is converted into kinetic energy (1/2 mv^2). Setting these equal gives:
mgh = 1/2 mv^2
Solving for v gives:
v = sqrt(2gh)
where m is the mass of the ball (0.125 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (3.421 m). Plugging in these values gives:
v = sqrt(29.83.421) = 8.2 m/s
Step 3: Calculate the horizontal distance the ball travels after it is released. Since the ball is released at the top of the circle, it travels in a parabolic trajectory. The horizontal distance it travels before hitting the ground is given by the equation:
d = vt
where d is the distance, v is the horizontal velocity (which is equal to the velocity at the top of the circle since there are no horizontal forces), and t is the time it takes for the ball to hit the ground. The time can be found using the equation:
t = sqrt(2h/g)
Plugging in the values gives:
t = sqrt(2*3.421/9.8) = 0.84 s
and
d = 8.2*0.84 = 6.9 m
So, the ball hits the floor 6.9 m to the right of the release point.
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