Question

The acceleration at which the block falls is uniform . That allows us to use kinematic equations along the y-direction to determine an equation for time.

Then, solving for time in terms of h and v f, we get that

also .This problem also allows us to investigate the angular portion of the spool, which is spinning without slipping as the block is accelerating downwards.

What is an expression for the angular displacement the spool goes through while the block is in motion?

(F1) θ = 2 g h 2 R v f 2

(F2) θ = h R

(F3) θ = 1 4

(F4)

INSTRUCTIONS: Answer the subsequent set of questions to solve the "Goal" of this problem. We recommend you write out your answer to compare to the particular questions as you go along.

SETUP: As shown in the figure, a box of mass m is attached to a light string that is wrapped around a cylindrical spool of radius R and mass M. The spool is suspended from the ceiling. Then, the box is released from rest a distance h above the floor.

FIGURE:

GOAL: Determine an expression for the amount of time it takes for the box to reach the floor
PART A - Totalinitial energy

We will start with energy conservation to solve this problem. Take the floor as our vertical "zero".

After removing all components that either cancel or equal "0", what does the total INITIAL energy look like?

(A1) K i only

(A2) K r o t i only

(A3) U g r a v i only

(A4) W o t h e ronly

(A5) K i + K r o t i

(A6) K i + K r o t i + U g r a v i

(A7) K i + K r o t i + U g r a v i + W o t h e r

(A8) K i + U g r a v i

Group of answer choices

A1

A2

A3

A4

A5

A6

A7

A8

PART B - Total final energy

Following the same outline as previous, after removing all components that do not change or that equal "0", what does the total FINAL energy look like?

(B1) K f only

(B2) K r o t f only

(B3) U g r a v f only

(B4) W o t h e r only

(B5) K f + K r o t f

(B6) K f + K r o t f + U g r a v f

(B7) K f + K r o t f + U g r a v f + W o t h e r

The acceleration at which the block falls is uniform (can you think of why?). That allows us to use kinematic equations along the y-direction to determine an equation for time.

Then, solving for time in terms of h and v f, we get that

Answer 1

SETUP: a box of mass m is attached to a light string that is wrapped around a cylindrical spool of radius R and mass M. The spool is suspended from the ceiling. Then, the box is released from rest a distance h above the floor.

I got an idea how the setup looks like.

There is a box hanging from a spool. As the spool rotates, the box goes down.

_____________

Part A

As we interested in box as of now

initial energy is gravitational energy only.

so,

correct answer - U_{gravitational}

_________________

Part B

now, the box has reached h = 0

so,

U ( grav) = 0

so,

final energy is just kinetic energy of box and rotational kinetic energy of spool

so,

correct option - K_f + K_rot, _f

______________________________

Part C

To solve for time in terms of vf and h

the linear acceleration of block is

a = vf² / 2h ( assuming it starts from rest - it is actually)

so,

vf = vi + at

t = 2h / v_f

_________________________________________________

angular displacement the spool goes through while the block is in motion

= 1/2 * * t²

= 1/2 * a/R * 4h² / vf²

= h/R