A stunt man whose mass is 70 kg swings from the end of a 4.4 m long rope along the arc of a vertical circle. Assuming that he starts from rest when the rope is horizontal, find the tensions in the rope that are required to make him follow his circular path at each of the following points.
(a) at the beginning of his motion
(b) at a height of 1.5 m above the bottom of the circular arc
(c) at the bottom of the arc
Thank you
To find the tensions in the rope at different points along the stuntman's motion, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration. We'll need to consider different forces acting on the stuntman at each point.
(a) At the beginning of his motion:
When the rope is horizontal, the stuntman is undergoing circular motion, which requires a centripetal force to keep him moving in a circle. At the beginning of his motion, when the rope is horizontal, the only force acting on him is the tension in the rope.
Taking the upward direction as positive, the net force equation can be written as:
Tension - Weight = Mass x Centripetal acceleration
At the beginning of his motion, the centripetal acceleration is zero, and the weight of the stuntman can be calculated as Weight = mass x gravity.
Therefore, the equation becomes:
Tension - (mass x gravity) = 0
Now, we can substitute the given values:
Tension - (70 kg x 9.8 m/s²) = 0
Solving for tension, we have:
Tension = 70 kg x 9.8 m/s²
Therefore, the tension in the rope at the beginning of his motion is 686 N (Newtons).
(b) At a height of 1.5 m above the bottom of the circular arc:
At this point, the stuntman is at a height of 1.5 m above the bottom of the circular arc. To find the tension, we'll consider the forces acting on him, which include the tension in the rope and his weight.
Taking the upward direction as positive, the net force equation can be written as:
Tension - Weight = Mass x Centripetal acceleration
At this point, the centripetal acceleration can be calculated using the radius of the circular arc and the speed of the stuntman. However, since the speed is not given, we need to use the conservation of energy to find it.
The conservation of energy equation is:
Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy
At the beginning of his motion (when the rope is horizontal), the potential energy is zero and the kinetic energy is also zero.
At a height of 1.5 m above the bottom of the circular arc, the potential energy is given by:
Potential Energy = mass x gravity x height
The final kinetic energy can be calculated using the formula:
Kinetic Energy = (1/2) x mass x velocity²
Since the stuntman starts from rest, the final kinetic energy is zero.
Therefore, the conservation of energy equation becomes:
0 + 0 = (mass x gravity x height) + 0.5 x mass x velocity²
We can solve this equation for velocity.
Once we have the velocity, we can calculate the centripetal acceleration using the formula:
Centripetal acceleration = (velocity²) / radius
Finally, we can substitute the values of mass, gravity, height, and radius into the net force equation:
Tension - (mass x gravity) = mass x (velocity² / radius)
Solving for tension will give us the answer.
(c) At the bottom of the arc:
At the bottom of the circular arc, the tension in the rope and the weight of the stuntman both act in the same direction, so the net force equation becomes:
Tension + Weight = Mass x Centripetal acceleration
Substituting the values of mass, gravity, and radius into the equation, we can solve for tension.
Please note that the calculation may change depending on additional information provided such as the speed of the stuntman or the height of the circular arc.