Gravitational Forces and Torque Reviews

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SUMMARY

This discussion focuses on the application of gravitational forces and torque in various physics problems. Key calculations include determining horizontal forces in static equilibrium, potential energy changes on an inclined plane, and angular momentum from applied torque. Specific examples include a trapeze artist's force calculation yielding 462N, potential energy for a crate on a ramp resulting in 200J, and a bucket's rotational kinetic energy calculated at 400 Joules. The participants clarify misconceptions regarding height in potential energy formulas and the relationship between applied forces and friction.

PREREQUISITES
  • Understanding of static equilibrium and forces
  • Knowledge of potential energy and its calculation
  • Familiarity with torque and angular momentum concepts
  • Ability to analyze inclined planes and frictional forces
NEXT STEPS
  • Study the principles of static equilibrium in physics
  • Learn about potential energy calculations on inclined planes
  • Explore torque and angular momentum relationships in rotational dynamics
  • Investigate the effects of friction on motion and energy transfer
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Students and educators in physics, particularly those focusing on mechanics, as well as anyone involved in solving real-world problems related to forces, energy, and motion.

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Here are some problems that I have figured out. Am I doing them right?

47.) A trapeze artist weighs 800N. The artist is momentarily held to one sid of a swing by a partner so that both of the swing ropes are at an angle of 30.0 degrees with the vertical. In such a condition of static equilibrium, what is the horizontal force being applied by the partner?

tan30 = opp/800. F = 800tan30 = 462N. Is this right?

65.) A 40.0N crate is pulled uip a 5.0m inclined plane at a constant velocity. If the plane is inclined at an angle of 37 degrees to the horizontal and there is a constant force of friction of 10.0 N between the crate and the surface, what is the net gain in potential energy by the crate?

Potential Energy = mgh. (40*5)-(40*0) = 200 J. However, the choices for this question are 120J, -120J, 210J, -210J. What am I doing wrong here?

68.) A child riding a bicycle has a total mass of 40.0kg. The child approaches the top of the hill that is 10.0m high and 100.0m long at 5.0 m/s. If the force of friction between the bicycle and the hill is 20.0N, what is the child's velocity at the bottom of the hill?

Force of gravity causes a force of 392.4N, 392.4cos84.26 = the applied force results in 39.25N. Net force = 19.25N, which gives a final velocity of 11 m/s. Is this correct?

85.) To warm up before a game, a baseball pitcher tosses a 0.15kg ball by rotating his forearm, which is 0.32m in length, to accelerate the ball. The ball starts at rest and is thrown at a speed of 12 m/'s in 0.40s. While the ball is in the pitcher's hand, what torque is applied to the ball to produce the angular acceleration?

I calculated that the acceleration created from his arm = 30 rad/sec^2. Torque = moment of inertia * ang. acceleration. I am missing my other paper of notes but my final calculations yield an answer of 1.4 N-m. Is this correct?

96.) A grinding wheel with a moment of inertia of 2.0kg-m^2 is initially at rest. What angular momentum will the wheel have 10.0 s after a 2.5 N-m torque is applied to it?

I calculated angular acceleration to = 1.25 rad/sec^2. Multiplied by 10 secs, that gives an angular speed of 12.5 rad/sec. Multiplied by the moment of inertia gives 25 kg-m/s^2 of Angular momentum. Is this correct?

100.) A bucket filled with water has a mass of 23kg and is attached to a rope that is woundwith a crank arounda 0.05m radius cylinder at the top of a well. The moment of inertia of the cylinder and crank is 0.12kg-m^2. The bucket and water are first raised to the top of the well and then released to fall back into the well. What is the rotational kinetic energy of the cylinder and crank at the instant the bucket is moving at a speed of 7.9 m/s?

I calculated that it took .81 s for the bucket to fall to a speed of 7.9 m/s. I calculated that the angular speed was 81.65 rad/sec. KE = (1/2)Iw^2 = (1/2)(.12)(81.65^2) = 400 Joules. Is this correct?
 
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47 is correct
65 - the h in the potential energy formula represents the HEIGHT change
what is 5m? Is it the HEIGHT of the ramp or the length of the ramp? Draw a diagram if you are still confused

you're going to have to be more speific on 68 coz I am not sure what ur thought pattern was there
 
Last edited:
65 - the h in the potential energy formula represents the HEIGHT change
what is 5m? Is it the HEIGHT of the ramp or the length of the ramp? Draw a diagram if you are still confused

If the hypotenuse was 5m, then the answer would be 120J. If the length of the base was 5.0m, the answer would be 150J, no a possible choice. If the Height was 5.0m, the answer would be 200J, nota possible choice.

68.) The angle between the force of gravity and the surface the child is riding on create an angle that can be used to determine the "applied force", right? Then, the net force, applied - frictional, is equal to mass * acceleration. Final Velocity^2 = Initial Velocity^2 + 2(acceleration)(distance). Choices are a.) 5.0 m/s b.) 10.0m/s c.) 11 m/s.
Cant be A as gravity would only increase the final velocity.
 
on 100, if the bucket is going at 7.9 m/s, then w=7.9/r, with r=0.05m?

I think your angular velocity is suspect...
 
100.) w = 81.65 radians / sec. Its translation motion, in free fall due to gravity is falling at a rate of 7.9 meters per sec.
 

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