Anahid nos muestra una excelente sesión de estadística, donde trataron el tema del error estándar.…

# Caída libre y Tensión: Completa Tutoría de física con estudiante

**Student**

3. A grinding stone in the shape of a solid disk with diameter .52 m and a mass of 50 kg is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 N, And the grinding stone comes to rest in 7.5 seconds. Find the coefficient of friction. 4. A 15 kg bucket of water is suspended by a rope wrapped around a windlass, that is solid cylinder .3m in diameter with mass 12kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10m to the water. You can ignore the weight of the rope. A. What is the tension in the rope while the bucket is falling? B.With what speed does the bucket strike the water? C. What is the time of fall? D. While the bucket is falling, what is the force exerted on the cylinder by the axle?

**Tutor**

Hi! Welcome!!

**Student**

hey! its nice to meet you. I just have a couple of problem i need help on. thanks for your help!

**Tutor**

Nice to meet you too!!

ok

let me see problem 3 please

one sec please

**Student**

ok

**Tutor**

ok

let me draw a diagram please

**Student**

ok!

**Tutor**

ok

Can you see the diagram?

**Student**

yup!

**Tutor**

the ax exerts a normal force of Fn = 160 N

**Student**

yup

**Tutor**

ok good

Can you remember the equation for friction force??

**Student**

right?

**Tutor**

correct!!

**Student**

Fk=Fnx(Mue sub k)

**Tutor**

we need to find mu sub k

right?

**Student**

yup!

but we don’t know Fk so, we need to find that right?

**Tutor**

exactly

**Student**

so can i make a force diagram?

**Tutor**

sure

**Student**

right?

**Tutor**

we can assume the disk is horizontal

we don’t need to consider the weight Fg

ok?

**Student**

right. but i just put that to balance it out.

**Tutor**

ok

**Student**

would Fa= (Omega) r?

**Tutor**

what is Fa ?

**Student**

Fa= force applied

**Tutor**

the only force applied is Fn

**Student**

i dont know that force, but there must be a counter acting force, to make Fk, and also that means there must be a constant velocity or accleration

**Tutor**

and it makes the disk stop

We can see that the only force applied is Fn

ok?

**Student**

ok

**Tutor**

due to this force, a friction force Fk = u Fn appears

and this friction force Fk makes the disk stop

right?

**Student**

right

**Tutor**

ok

now we can find the angular acceleration of the disk

**Student**

alpha= (omega) r right?

**Tutor**

correct!!

sorry

let me change something

**Student**

so i get 23.143?

**Tutor**

V = w r

and

a = alpha r

**Student**

oh.

**Tutor**

Can you see the eqs?

V = w r

a = alpha r

**Student**

yup!

**Tutor**

ok

to find alpha we need to remember a kinematics equation

we know the initial angular speed

the time t

and the final speed

correct!!

correct!!

wf = alpha t + wo

so what is alpha?

**Student**

so Alpha=11.87

–

-11.87 rads/sec^2

**Tutor**

what is the unit?

it is better to work with rad/s

for angular speed

ok?

**Student**

ok

oh my bad, yeah thats what i got for Wo

**Tutor**

perfect!!

**Student**

then i divided it by time

to get my alpha

**Tutor**

alpha = -11.87 rad/s^2

**Student**

yes, that is what i got!

**Tutor**

good

now

we need to remember the Newton’s second law for rotations

for linear motion we know Fnet = ma

right?

**Student**

yup!

**Tutor**

and for rotations we have a very similar expression

Torque(net) = I alpha

let me write it on the board please

Can you see the eq on the board?

**Student**

yes

**Tutor**

is this expression familiar to you?

Tnet = net torque

I = moment of inertia

alpha = angular acceleration

**Student**

right

**Tutor**

ok good

what is I for a solid disk?

**Student**

um. 1/2MR^2?

**Tutor**

perfect!!!!

so let’s find I for the grinding stone

I = ?

**Student**

1.69?

**Tutor**

yes!!

I = kg m^2

**Student**

then i get a torque of -20.1?

**Tutor**

yes!!

Tnet = 20.06 N m

Can you remember the definition for torque?

the torque applied on the disk is produced by the friction force

right?

**Student**

uh yeah it makes sense, since the torque goes against omega

**Tutor**

yes!!

and torque = F d

right?

let me write it on the board

please

**Student**

oh right, yeah that makes sense

**Tutor**

ok good

so we have the eq in magnitude

**Student**

so Mu sub k =.483?

**Tutor**

20.06 Nm = u Fn (0.26)

yes!!

0.48

Great job on that one! Nicely done.

Do you have any questions about what we just did?

**Student**

no, i think you explained it well enough, i believe i understand the problems a lot better, thank you!

**Tutor**

you are welcome

now problem 4

Can you give me a moment while I take a look at problem 4??

**Student**

ok

**Tutor**

ok

let me draw a diagram please

on a new board

**Student**

k

**Tutor**

ok can you see the diagram?

**Student**

yup!

**Tutor**

ok good

what forces are acting on the bucket?

**Student**

gravity, and the rope?

**Tutor**

good

gravity and tension

ok

for part A we have to find the tension

right?

**Student**

yup

so Ft=Mbg?

so Ft=147N?

**Tutor**

Fg = mb g

Fg = 15 kg x 9.8 m/s^2

Fg = 147 N

right?

**Student**

oh right

**Tutor**

but it is not T

the bucket is falling

with a certain acceleration a

right?

**Student**

yes

**Tutor**

and for the bucket we have Fnet = m a

or

Fg – T = mb a

right?

**Student**

yes

that makes sense

**Tutor**

ok

a is the acceleration of the bucket

and it is also the tangential acceleration of teh cylinder

right?

do you agree??

**Student**

yes, that would make sense

**Tutor**

ok

let me write one more eq please

on the white board

**Student**

k

**Tutor**

what is the moment of inertia of a solid cylinder?

**Student**

i believe the same as the disk problem

**Tutor**

good

1/2 m r^2

**Student**

so .135 Alpha= Tr

**Tutor**

ok good

and alpha = a/r

right?

**Student**

right

**Tutor**

ok good

I labeled equations 1 and 2

can you see it??

**Student**

yup

**Tutor**

we can take these two eqs

and form a system of equations

right?

**Student**

yup

**Tutor**

we have two eqs

and two unknown

variables

a and T

**Student**

so should we plug in for T or a?

**Tutor**

we need both

**Student**

i think a right?

**Tutor**

yes

any is ok

**Student**

so i simplified it. and got a=T/6

is that right?>

**Tutor**

perfect!!

now let’s replace a in eq 1

**Student**

so i get T=42?

**Tutor**

perefct!!

T = 42 N

good

is it clear so far?

**Student**

yeah, things are clearing up, thanks a bunch!

**Tutor**

note that we are using the same equations we used in the previous problem

Tnet = I alpha

T = F d

and

a = alpha r

right?

**Student**

yeah!

**Tutor**

and the Newton’s second law

Fnet = ma

ok good

let’s see part B

B. With what speed does the bucket strike the water?

we can find a now

right?

what is a?

**Student**

7m/s^2

**Tutor**

yes a= 7 m/s^2

now we have a kinematics problem

the bucket starts from rest

and falls 10 m

what is the final velocity?

let me see

perfect!!

Vf = 11.83 m/s

**Student**

and that means the time to hit the water is 1.69?

**Tutor**

yes!!

t = 1.69 s

**Student**

ok part D. this one is more complicated

would it be 147N?

**Tutor**

it would be the same magnitude as tension

right?

the cylinder is rotating

but it is in translational equilibrium

**Student**

so it would not be 42N?

**Tutor**

T = 42 N

**Student**

right. but that wouldnt be the force on the cylinder?

**Tutor**

yes

it would be the force on teh cylinder applied by the axle

**Student**

ok so how do we find this force?

**Tutor**

let me draw a diagram please

Can you see the diagram?

the force F is exerted by teh axle

on the cylinder

right?

**Student**

yeah that makes sense

**Tutor**

the cylinder is in translational equilibrium

so the equilibrium conditions must be verified

Sum Fx = 0

Sum Fy =0

Sum Fy = F – T = 0

right?

**Student**

right that makes sense

and there is no force in the X direction right?

**Tutor**

exactly

so F = T

**Student**

so F=42?

**Tutor**

good

**Student**

so it is 42?

**Tutor**

exactly

You got it!

**Student**

haha. thanks.

**Tutor**

Do you have any questions about what we just did?

**Student**

no i think i got the jist of it. thank you so much!

**Tutor**

you are welcome

Thanks.

**Student**

have a good day!

**Colaborador: Luis Rodríguez, Tutor Boliviano de Física**