Question

Solid circular disk of radius (0.5 m) rotates around its center of mass under effect of tangential force (F) resulted from pulling a robe coiled around the disk.
The angular displacement is given according to the following relation:
θ(t)=5t^3+8t^2+4t+10

Given that the torque effecting the disk at t = 2s
is 11.4 N.m
Find the following:
A) Mass of the disk : (Ok)

a) 1.2 kg b) 0.15 kg c) 0.3kg d) 0.78 kg

B) The force (F) at t = 3s:

a)15.9 N b) 31.8 N c) 63.6 N d) 35 N

C) Average angular acceleration during t = 4, t =0

D) Length of pulled robe during t = 4, t= 0:

Answer 1

I manged to solve the rest, but got different answers than the aimed. ( in choices, but not correct).

Answer 2

for which q u didn't get the answer?

Answer 3

B, C, D :(

Answer 4

first B

Answer 5

how can u relate torque and force?

Answer 6

M = 1.2 kg

Answer 7

I got the force as a function of time. and let t = 3 s

Answer 8

\[\theta (t) = 5t^{3} + 8 t^{2} + 4t + 10\]

Answer 9

i was about to ask that

Answer 10

My bad.

Answer 11

u got the force as a function of time?

Answer 12

I did.

Answer 13

then what is the problem?

Answer 14

My answer differs from the book.

Answer 15

Write your solution and then we can compare it to mine.

Answer 16

sorry i m not sure..pls check with someone else

Answer 17

@dan815

Answer 18

well..did u differentiate theta to find the angular acceleration and find it 2s?

Answer 19

yeah

Answer 20

then did u divide by r ?

Answer 21

to get the linear acceleration..??

Answer 22

\[\theta (t) = 5t^3 + 8t^2 + 4t +10\]
\[\omega (t) = 15t^2 + 16t +4\]
\[\omega ' (t) = 30t + 16\]

\[F = ma_{t}\]

\[a_{t} = \omega r \]

\[F = mr \omega = 0.6 (30t + 16) = 18t+ 9.6\]

\[F(t) = 18t + 9.6\]

Answer 23

\[F(3) = 18 * 3 + 9.6 = 63.6 N\]

Answer 24

Included in the choices, but not the correct answer.

Answer 25

even i did the same way..so i think ur answer is right

Answer 26

but i took the mass u gave is that correct?

Answer 27

??

Answer 28

Check yourself.

Answer 29

The book has full solution, not just final answer.

Answer 30

I recalculated all his steps and he keep getting 31.8 N

which is 0.5 F(3) mine.

Answer 31

umm what ddid u get for the mass?

Answer 32

1.2 kg

Answer 33

ya i checked it..its 1.2kg

Answer 34

where has the book deviated from the above steps?

Answer 35

Book solution for (B):

\[\tau = Fr\]

\[\tau = I \omega '\]

\[\omega ' (3) = 30(3) + 16 = 106 rad/s^2\]

\[\tau (3) = 0.15 * 106 = 15.9 N.m\]

\[F = \frac{ \tau }{ r } = \frac{ 15.9 }{ 0.5 } = 31.8 N\]

Answer 36

oh ur off by a factor of 2 hmm

Answer 37

or r^2

Answer 38

if it was

F = mr^2 omega

Answer 39

hey u have substituted w' in place of w

Answer 40

Typo it must be

a = w' r

F = w' mr

Answer 41

oh ya..

Answer 42

but does this condition hold true always?

Answer 43

\[s = \theta r \]

\[\frac{ d }{ dt }\]


\[v = \omega r\]


\[\frac{ d }{ dt }\]

\[a_{t} = \omega ' r\]

Answer 44

In the proof, only r (the radius) was constant. and it is.

Answer 45

No i am asking like..only in pure rolling a=w'/r is valid right?

Answer 46

Nah, it always hold true.

as

S = arc length
θ = angular displacement
r = raduis

S = θr

Answer 47

the one of pure rolling you're talking about is

M = a_{r} * N ?

Answer 48

Later in the question, it was used to find D

Answer 49

i didn't get ur notation..a_{r}??

Answer 50

\[M = a_{r} N\]

Answer 51

what does N stand for?

Answer 52

Normal force.

Answer 53

Reaction of the surface upon a circular tire.

Answer 54

ok and M?

Answer 55

The moment rotating the body.

Answer 56

its seems as though both methods are correct...but the answers come out to be different..
if i get any idea i'll tell u

Answer 57

This is a rotational dynamics question with non-uniform angular acceleration.
Check my working!

Answer 58

@Farcher , Nice, your steps are totally correct. Can you spot the mistake in my own solution ?

Answer 59

ya @Farcher great job! but why can't we do \[a=\alpha*r\] ?

Answer 60

eventhough the acceleration is non-uniform we are finding its value only at a particular instant..right?

Answer 61

a = alpha r is correct but it is not constant acceleration.

Answer 62

Well, I found the force as a function of time, the solution is above check it.

Answer 63

Then\[\tau = I_c \alpha \Rightarrow Fr = \frac 1 2 m r^2 \alpha \Rightarrow F = \ \frac 12 m a\]

Answer 64

Nice proof, but why isn't the force F = ma as usual ?

Answer 65

i think Farcher is right..here we can't straight away apply F=ma coz the object is under rotational motion..inertia is the rotational equivalent of mass

Answer 66

Because it is not a point mass. It is a disc which is a lot of point masses distributed over a volume.
There must actually be another equal and opposite force acting at the axle so that the disc just undergoes rotation and its centre of mas does not move.

Answer 67

Thanks for the explanation..

Answer 68

I now got it , F = ma , resultant of forces = ma ( newton law) i used it as the sole force.

Answer 69

Thanks Farcher.