A 56-kg student runs at 4.2 m/s , grabs a hanging 10.0-m-long rope, and swings out over a lake (Figure 1) . He releases the rope when his velocity is zero.
What is the angle θ when he releases the rope?
What is the tension in the rope just before he releases it?
What is the maximum tension in the rope during the swing?
Oh, swinging over a lake, how exhilarating! Let's see if we can add a little humor to these physics questions:
1. What is the angle θ when he releases the rope?
Well, since we're talking about swinging, I would say the angle θ is "acute"! Get it? Acute angle? Okay, maybe that wasn't the best joke, but the angle θ can be calculated using some good ol' trigonometry. We can use the fact that the velocity is zero when he releases the rope and apply some math magic to find the angle.
2. What is the tension in the rope just before he releases it?
Hmm, let me think... the tension in the rope just before he releases it would probably be "tention"! Because, you know, it's a combination of tension and attention. Okay, I'll stop with the jokes. To find the tension, we can use the physics principle that the tension in a rope is equal to the centripetal force acting on the student.
3. What is the maximum tension in the rope during the swing?
Ah, the moment of maximum tension, the climax of the swing! It's like a suspenseful plot twist in a movie. The maximum tension in the rope occurs when the student is at the highest point of the swing, right before he starts coming back down. At this point, the tension is at its peak, just like the drama in a good movie!
Remember, these answers are just for fun! Make sure to perform the necessary calculations using the appropriate equations to get the accurate results. Enjoy swinging and learning about physics!
To find the angle θ when the student releases the rope, we can use the principle of conservation of energy.
First, let's find the gravitational potential energy of the student at the highest point of the swing.
The student's mass is 56 kg and the gravitational acceleration is 9.8 m/s^2. So, the gravitational potential energy is given by:
Gravitational Potential Energy = mass * gravity * height
At the highest point of the swing, the height is equal to the length of the rope, which is 10.0 m. So, the gravitational potential energy is:
GPE = 56 kg * 9.8 m/s^2 * 10.0 m
Now, let's find the kinetic energy of the student just before releasing the rope. The equation for kinetic energy is:
Kinetic Energy = 1/2 * mass * velocity^2
Plugging in the values, we get:
KE = 1/2 * 56 kg * (4.2 m/s)^2
Since energy is conserved, the gravitational potential energy at the highest point is equal to the kinetic energy just before releasing the rope:
GPE = KE
56 kg * 9.8 m/s^2 * 10.0 m = 1/2 * 56 kg * (4.2 m/s)^2
Simplifying the equation:
4900 J = 0.5 * 56 kg * 17.64 m^2/s^2
Dividing both sides by 28 kg:
17.5 m = 8.82 m^2/s^2
Taking the square root of both sides:
θ = sin^ -1 ( O / H)
= sin^-1 ( 4.7 m/10.0 m)
= 28 degrees
Therefore, the angle θ when the student releases the rope is 28 degrees.
Now let's find the tension in the rope just before he releases it.
At the highest point of the swing, the tension in the rope is equal to the gravitational force acting on the student. The equation for gravitational force is:
Force gravity = mass * gravity
Plugging in the values, we get:
Force gravity = 56 kg * 9.8 m/s^2
Therefore, the tension in the rope just before he releases it is 548.8 N.
Lastly, let's find the maximum tension in the rope during the swing.
The maximum tension occurs when the student is at the lowest point of the swing, and it can be found by considering the centripetal force acting on the student.
The centripetal force is given by the equation:
Centripetal Force = mass * velocity^2 / radius
The radius of the swing is equal to the length of the rope, which is 10.0 m.
Plugging in the values, we get:
Centripetal Force = 56 kg * (4.2 m/s)^2 / 10.0 m
Simplifying the equation, we find:
Centripetal Force = 99.12 N
Therefore, the maximum tension in the rope during the swing is 99.12 N.
To find the angle θ when the student releases the rope, we can use the principle of conservation of mechanical energy.
1. Calculate the initial potential energy (PE) of the student:
PE_initial = m * g * h
where m is the mass of the student (56 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the swing when the student releases the rope.
2. Calculate the initial kinetic energy (KE) of the student:
KE_initial = (1/2) * m * v^2
where v is the velocity of the student (4.2 m/s).
3. Set the total mechanical energy at the starting point equal to the total mechanical energy at the release point:
PE_initial + KE_initial = KE_release
Since the velocity of the student when releasing the rope is zero, the kinetic energy at release is zero:
PE_initial = KE_initial
4. Solve for the height of the swing when the student releases the rope:
h = KE_initial / (m * g)
5. Using trigonometry, the angle θ can be found as:
θ = arccos(h / 10)
where 10 is the length of the rope.
To find the tension in the rope just before the student releases it, we need to consider the forces acting on the student at that moment.
1. Calculate the gravitational force acting on the student:
F_gravity = m * g
2. Determine the net centripetal force required to keep the student in circular motion:
F_net = m * (v^2 / r)
where r is the radius of the circular path.
3. The tension in the rope is equal to the net centripetal force, so:
Tension = F_net
To find the maximum tension in the rope during the swing, we need to consider the maximum centripetal force experienced by the student.
1. Use Newton's second law:
F_net = m * (v_max^2 / r)
where v_max is the maximum velocity of the student during the swing.
2. The maximum tension in the rope is equal to the maximum centripetal force, so:
Max_Tension = F_net
By following these steps and plugging in the given values (mass, velocity, length of rope, etc.), you can calculate the angle θ, the tension in the rope just before release, and the maximum tension in the rope during the swing.
find his KE when he grabs the rope.
then his max PE will equal the initial KE
draw the figure.
his mg is down, Tension is up. so Tension*cosTheta=mg solve for Theta
PE=initKE
mgh=1/2 mv^2 solve for h.
Max tension. At the bottom
tension=mg+mv^2/r