Why Are These Physics Problems So Challenging?

[tandoorichicken]

I hate to be taking up so much board space today, but all my problems are so freakin difficult.

1. a) show that the kinetic energy of a satellite in circular orbit r about the Earth is one half the magnitude of its gravitational potential energy.
b) what is its total mechanical energy?

2. Okay, I have to describe the picture in the book for this one: There is a wheel of r = 0.38m and m = 1.3kg and attached to that wheel from a cord is a 0.70kg block that is 1.2m off the ground.
a} If the block is released from rest, what speed will it have just before it hits the floor if there is no friction at the wheel's axis?
b) If there is friction on the axis, what is the frictional torque if the speed just before the block hits the floor is half of the speed it would have without friction.

 

[himanshu121]

Write the formula for K.E ,Centripetal force Equation and P.E u will get the result Mechanical Energy= P.E+K.E

 

[M98Ranger]

First of all, it's completely understandable that you are struggling with these difficult physics problems. Physics can be a challenging subject, and it's important to take the time to work through them and understand the concepts rather than rushing through them.

In regards to the first problem, we can use the equation for gravitational potential energy, U = -GmM/r, where G is the gravitational constant, m is the mass of the satellite, M is the mass of the Earth, and r is the distance between them. We can also use the equation for kinetic energy, K = 1/2mv^2, where v is the velocity of the satellite.

To show that the kinetic energy is one half the magnitude of the gravitational potential energy, we can set the two equations equal to each other and solve for v. This will show us the velocity of the satellite in circular orbit.

For the total mechanical energy, we can simply add the kinetic and potential energies together, since they are both forms of energy. This will give us the total energy of the satellite in circular orbit.

Moving on to the second problem, we can use the equation for rotational kinetic energy, K = 1/2Iω^2, where I is the moment of inertia and ω is the angular velocity. First, we need to find the moment of inertia for the wheel, which is given by I = 1/2mr^2.

a) If we assume that the block is released from rest, we can use the conservation of energy to find the final velocity of the block just before it hits the ground. This will be equal to the velocity of the block at the bottom of the wheel, since there is no friction at the axis.

b) However, if there is friction at the axis, we need to consider the work done by friction, which is equal to the frictional torque multiplied by the angle through which it is applied. We can use this to find the final velocity of the block just before it hits the ground, given that it is half of the velocity without friction.

I hope this helps you in your understanding of these difficult physics problems. Keep persevering and don't be afraid to ask for help when needed. Good luck!