A student is swinging a bowling ball (mass of 25.0 kg) in circles that is connected at the end of a massless rope (length of .75 m) The student lets go of the rope, and the rope makes an angle 63.5 degrees with the vertical. What is the Tmax and vmax when the student lets go of the rope?
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
The tension force (T) in the rope is responsible for providing the centripetal force that keeps the bowling ball moving in a circular path. When the student lets go of the rope, the centripetal force is no longer provided, and the bowling ball will move off in a straight line tangential to the circle.
To solve for the tension force (T), we can use the centripetal force equation:
F = ma = mv²/r
where F is the centripetal force, m is the mass of the bowling ball, v is its velocity, and r is the radius of the circular path.
The centripetal force is provided by the tension force, so we can substitute T for F:
T = mv²/r
We can solve for v by using the conservation of energy equation:
K + U = constant
where K is the kinetic energy of the bowling ball, and U is its potential energy.
At the highest point of the circle, the potential energy is at a maximum, and the kinetic energy is at a minimum. At the moment the student lets go of the rope, all of the potential energy is converted to kinetic energy, so we can set U = 0 and solve for K:
K = mv²/2
Setting K equal to its maximum value (the initial potential energy), we can solve for v:
mgh = mv²/2
v² = 2gh
where h is the height of the bowling ball above its lowest point at the moment the student lets go of the rope.
We can solve for h using trigonometry:
h = L(1 - cosθ)
where L is the length of the rope, and θ is the angle the rope makes with the vertical.
Substituting h and solving for v:
v² = 2gL(1 - cosθ)
v = sqrt(2gL(1 - cosθ))
Now we can substitute this expression for v into the equation for T and solve for Tmax, the maximum tension in the rope:
Tmax = mv²/r
Tmax = (m/sqrt(2gL(1 - cosθ)))²/r
Tmax = mg(1/(sinθ/2))
Substituting the given values, we get:
Tmax = (25.0 kg) * (9.81 m/s²) * (1/sin(63.5°/2))
Tmax = 674 N
Therefore, the maximum tension in the rope is 674 N when the student lets go of the rope.
To find vmax, we can substitute the given values into the expression we derived earlier:
vmax = sqrt(2gL(1 - cosθ))
vmax = sqrt(2 * 9.81 m/s² * 0.75 m * (1 - cos 63.5°))
vmax = 5.05 m/s
Therefore, the maximum velocity of the bowling ball is 5.05 m/s when the student lets go of the rop
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