Angular Speed of Pulley (Spring, Pulley, Ramp)

[Ticklez_Panda]

Homework Statement

---Find Angular Speed---

Homework Equations

I = 1/2 mR^²
Ug = mgdsinθ
Us = 1/2 kd^²
KE = 1/2 I w^²

The Attempt at a Solution

Moment of inertia of reel = 1/2mR²
Kinetic energy of the reel = 1/2 mv² = 1/2(1/2mR²) w²
KE = mgdsinθ (potential due to gravity) + 1/2kd² (spring potential)
w = sqrt(4(mgdsinθ + 1/2kd²) / mR²)
This answer is incorrect, I'm wondering where I went wrong. Please help if you are able to. Cheers!

 

[voko]

You are forgetting that the block will also have non-zero KE.

 

[Tanya Sharma]

Moment of inertia of reel = 1/2mR²
Kinetic energy of the reel = 1/2 mv² = 1/2(1/2mR²) w²
KE = mgdsinθ (potential due to gravity) + 1/2kd² (spring potential)
w = sqrt(4(mgdsinθ + 1/2kd²) / mR²)
This answer is incorrect, I'm wondering where I went wrong. Please help if you are able to. Cheers!

You have not considered the kinetic energy of the block.

There are two types of kinetic energies to be taken into account.

1) Rotational kinetic energy of the reel given by (1/2)Iω²
2) Translational kinetic energy of the block given by (1/2)mv²

 

[Ticklez_Panda]

You have not considered the kinetic energy of the block.

There are two types of kinetic energies to be taken into account.

1) Rotational kinetic energy of the reel given by (1/2)Iω²
2) Translational kinetic energy of the block given by (1/2)mv²

Kinetic energy of the block must be equal to the spring energy? I assume this since the block will fall with potential due to gravity, during this fall it will gain kinetic energy and it is the accumulated energy that will transfer into the spring. Which means that 1/2mv^2 = 1/2kd^2.

So it would follow that w = sqrt(4(mgdsinθ + 1/2kd2 - 1/2mv^2) / mR2)

 

[Tanya Sharma]

Kinetic energy of the block must be equal to the spring energy? I assume this since the block will fall with potential due to gravity, during this fall it will gain kinetic energy and it is the accumulated energy that will transfer into the spring. Which means that 1/2mv^2 = 1/2kd^2.

So it would follow that w = sqrt(4(mgdsinθ + 1/2kd2 - 1/2mv^2) / mR2)

This is incorrect understanding...

Start afresh...

Initially,when the block is 'd' distance up the incline what is the total mechanical energy of the system ?

 

[Ticklez_Panda]

This is incorrect understanding...

Start afresh...

Initially,when the block is 'd' distance up the incline what is the total mechanical energy of the system ?

Initially ΔEint= 0 because the spring is not compressed and the block is attached to the reel. When the reel is wound counterclockwise the block begins to rise, it gains potential energy due to gravity and the pulley experiences kinetic energy as it rises. I'm not quite sure where to go from here. I'm a bit confused that the pulley is what were are seeking the speed of. Generally it's been about the speed of the block so far in the course.

 

[voko]

Well, the problem is intentionally set up so as to confuse you. Instead of one potential energy, it has two. Instead of one kinetic energy, it has two. Instead of purely linear motion, it has linear and rotational, which are kinematically related.

Yet you can solve it exactly as the other you did. Find the full potential energy. The full kinetic energy. Their sum is constant.

 

[Tanya Sharma]

I will let voko guide the OP.