How to calculate K.E of this skater?

  • Thread starter davidraj
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In summary, the ice skater's angular speed changes from 15 rad/sec to approximately 26.5 rad/sec when she pulls her arms in, leading to a significant increase in kinetic energy due to conservation of angular momentum. In the second conversation, the bus will stop before reaching the deer due to the application of brakes. To calculate the velocity of the bus when it reaches the deer, the equation for work done is -µk*m*g*d, taking into account the potential energy at the top position. In the third conversation, the forces of lift and air resistance must be resolved to determine their horizontal and vertical components, and then Newton's second law can be used to find the values of F and R.
  • #1
davidraj
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An ice skater spins about a vertical axis at an angular speed of 15 rad/sec when her arms are outstretched. She then quickly pulls her arms into her sides in a very small time interval so that the frictional forces due to ice are negligible. Her initial moment of inertia about the axis of rotation is 1.72 kg m2 and her final moment of inertia is 0.61 kg m2. What is the
change in her angular speed? What is the change in her kinetic energy? Explain this change
in kinetic energy.
 
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  • #2
The angular momentum is conserved as there is no net external torque on the skater. Equate it for her final and initial states to find the new angular velocity.
 
  • #3
Thank you pi-bond...yep i got the idea and ill pick up with this...can you kindly help me with this question too...
A bus is moving downhill at a slope of 5°. At the moment when the speed of the bus is
30 km/h, the driver spots a deer 30m ahead. He applies the brakes and comes to a stop. The deer is paralyzed by fear and does not move. Will the bus stop before reaching it or will it hit the deer? Do relevant calculations and draw appropriate force diagram. Take the coefficient of kinetic friction to be µk = 0.26.

I calculated the velocity of the bus when it reaches the deer using the formula, Work done = 1/2 mv1 square - 1/2 mv2 square...where work done is given by -µ . m . g . d...the velocity of the bus comes around 27 km/h when it reaches the deer...is my equation correct? if not kindly suggest the method of solving this question...
 
  • #4
You also have to consider the potential energy which the bus has at it's top position; find the height between the two positions and then include the potential energy in your conservation equation.
 
  • #5
Thank you again Pi-Bond...can you kindly clear another doubt of mine...sorry for bugging...here's the question:

A plane is flying with a constant speed along a straight line at an angle of 60° with the
horizontal. The weight W of the plane is 90, 000 N and its engine provides a thrust T of
120, 000 N in the direction of flight. Two additional forces are exerted on the plane: the lift force F perpendicular to the plane’s wings, and the force R due to air resistance opposite to the direction of motion. Determine F and R.

Can you kindly tell me what steps and formulae should be used to calculate F and R?
 
  • #6
Can you draw a free body diagram showing all forces?
 
  • #7
plane.JPG


O is the startin point. OT gives the direction and magnitude of Thrust. RP stands for the resistance of air ( drag?) . FP is the Force (Lift) that acts perpendicular to the wings. Am sure I've got one or two of these wrong. Kindly correct.
 
  • #8
You forgot the weight, otherwise it is all right. Now resolve the forces along horizontal and vertical directions. Can you use the fact that the plane moves in a straight line with constant velocity to write an equality for the resolved forces?
 
  • #9
horizontal forces are the resistance of air and the thrust acting opposite to each other.
Vertical forces are the Force of Lift and the weight of the plane acting opposite to each other. Bur how do i derive an equation based on this statement?
 
  • #10
The forces due to thrust, air resistance and lift are not purely horizontal or vertical, as they are at an angle to the vertical and horizontal axes. You need to resolve them to these axes, and then use Newton's second law for horizontal and vertical forces.
 

1. How do I calculate the kinetic energy of a skater?

To calculate the kinetic energy of a skater, you will need to know the mass of the skater and their velocity. The formula for kinetic energy is KE = 1/2 * m * v^2, where m is the mass in kilograms and v is the velocity in meters per second.

2. What units are used to measure kinetic energy?

Kinetic energy is typically measured in joules (J), which is the standard unit for energy in the International System of Units (SI). However, in some cases, other units such as foot-pounds (ft-lb) or calories (cal) may be used.

3. How does the speed of the skater affect their kinetic energy?

The speed of the skater has a direct effect on their kinetic energy. As the speed of the skater increases, their kinetic energy also increases. This is because kinetic energy is directly proportional to the square of the velocity.

4. Can the kinetic energy of a skater ever be negative?

No, the kinetic energy of a skater cannot be negative. Kinetic energy is a scalar quantity, meaning it only has magnitude and no direction. In other words, it cannot have a negative value.

5. What other factors may affect the kinetic energy of a skater?

In addition to mass and velocity, other factors that may affect the kinetic energy of a skater include air resistance, friction, and the surface on which they are skating. These factors can either increase or decrease the kinetic energy of the skater.

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