# Rolling disc

1. Mar 28, 2005

### stunner5000pt

A thin string has been wound around a homogenous circular disc A in such a way that the disc can oroll on a horiz0ontal surface. The mass of the disc is M and its radius is R. The string goes around a small frictionless pulley C at a height 2Rabove the horizontal table B. To the free end of teh string is a bob of Mass M. The disc the pulley and the bob and the string are in the same vertical place.
The system begins to move under influence of gravity. The string is taut throughout and the mass of the string and pulley are negligible.
Also coefficient of friction mu is so large thte disc rolls without slipping.
Both are of same mass M so case of M doesnt matter
1. Determine the magnitude of the froce S which the string acts on the disc as long as teh systme moves under influence of gravity.

well the angle between the string and the disc is 90 so we dont haev to worry about the angle !

since there is no slipping $$a = \alpha R$$
the components of newton' second law
for the block mass : $$Ma = Mg - S$$
for the disc: $$Ma = F$$ let F be the force of friction
angular components of newton's second law
for the disc $$I \alpha = R (S - F)$$
$$I \frac{a}{R^2} = S - F$$
before i do any more simplification, i'd like to know if i have one everything right here ! Please do help!!

2. Determine the smallest value of mu that permits the disc to roll without slipping.
firs of all when something liek a disc is involved, there is only a point that is touching the surface. But do the same rules (like those of a block) apply here too?/

so then the force of friction would be $$F = \mu Mg$$
also $$Ma = F$$ soo, $$\mu Mg = Ma$$
$$\mu = \frac{a}{g}$$
is this the right direction??

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2. Mar 28, 2005

### stunner5000pt

all i really need is some corrections if there exist any,
i was wonering if for the linear components of the disc, the string force would also be included in which case things would change a bit, then...

3. Mar 28, 2005

### Gamma

Since F and S both act on the disc, isn't it

$$Ma = F + S$$

Other than this I didn't see any thing wrong with your solutions.

4. Mar 29, 2005

### stunner5000pt

so when theres torque involved you have to include forces are linear forces and angular forces?

5. Mar 29, 2005

### Gamma

Forces F and S causes a net torque resulting in the rotational motion of the disc. In other words, F and S causes the translational motion of the center of mass of the disc.

That's why I believe that net force (F + S) = Ma. Hope some one else will comment on this.

6. Mar 29, 2005

### Staff: Mentor

You are correct, Gamma.

Regarding part 2: First solve for the friction force F. The minimum value of $\mu$ will be given by setting $F = \mu Mg$.

7. Mar 29, 2005

### xanthym

From the problem statement:
{String Tension} = S
{Mass of Suspended Object} = M
{Weight of Suspended Object} = W = M*g
{Acceleration of Suspended Object} = A ::: <---- (different from "a")
{Disk Mass} = M
{Disk Radius} = R
{Disk Moment of Inertia} = I = (1/2)*M*R^2
{Disk Linear Acceleration} = a ::: <---- (different from "A")
{Disk Angular Acceleration} = α = a/R
{Coeff of Friction} = μ
{Friction Force on Disk} = Ffric = μ*M*g

For the Suspended Object (Linear Motion only):
{Net Force on Suspended Object} = {Susp Object Mass}*{Susp Object Acceleration} =
= M*A =
= {Suspended Object Weight} - {String Tension} =
= W - S =
= M*g - S =
::: ⇒ A = (M*g - S)/M ::: Eq #1

For Disk Linear Motion (of Disk Center of Mass):
{Net Force on Disk} = M*a =
= S - Ffric
::: ⇒ Ffric = S - M*a ::: Eq #2

For Disk Rotational Motion (about Disk Rotational Axis):
{Net Torque about Rotational Axis} = Iα = I*a/R = a*(1/2)*M*R^2/R = a*(1/2)*M*R =
= S*R + Ffric*R
::: ⇒ Ffric = (1/2)*M*a - S ::: Eq #3

Relationship between Suspended Object Acceleration ("A") and Disk Linear Acceleration ("a") due to String Unwinding (combination of motion of Disk Center of Mass and Disk Rotational Motion unwinding string):
A = a + R*α = a + (a) = 2*a ::: Eq #4

Combining Eq #1 & Eq #4, and solving for "a":
A = (M*g - S)/M ::: ⇒ 2*a = (M*g - S)/M
::: ⇒ a = (M*g - S)/(2*M) ::: Eq #5

Equating Eq #2 and Eq #3:
S - M*a = (1/2)*M*a - S
::: ⇒ S = (3/4)*M*a ::: Eq #6

Placing Eq #5 into Eq #6, we get:
S = (3/4)*M*{(M*g - S)/(2*M}
::: ⇒ {String Tension} = S = (3/11)*M*g ::: Eq #7

From Eq #2, Eq #5, and Eq #7, we have:
Ffric = μ*M*g =
= S - M*a =
= (3/11)*M*g - M*{(M*g - S)/(2*M} =
= (3/11)*M*g - (1/2)*M*g + (1/2)*S =
= (3/11)*M*g - (1/2)*M*g + (1/2)*{(3/11)*M*g} =
= (-1/11)*M*g
::: ⇒ {MINIMUM μ} = (1/11)

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Last edited: Mar 30, 2005
8. Mar 30, 2005

### Staff: Mentor

In addition to Gamma's comment, beware of using the same symbol (a) for the accelerations of the disk and the hanging mass. They are related, but not the same. (Find the acceleration constraint connecting the two.)

9. Mar 31, 2005

### stunner5000pt

A = a + R*α = a + (a) = 2*a
Why does this relaton hold true??

why is A = a + R*α??

also Ffric = (1/2)*M*a - S
implies that the friction and the string torques are in the same direction

since they both point clockwise right?/

Last edited: Mar 31, 2005
10. Mar 31, 2005

### Staff: Mentor

The acceleration of the falling mass equals the acceleration of the string itself. You can think of the string's acceleration as having two parts: the acceleration of the disk's center (a) and the acceleration of the unwrapping string with respect to the disk's center ($\alpha R = a$). The string's acceleratoin is the sum of those parts.

Since the friction and string tension both pull to the right (see Gamma's comment), I would write that torque-related equation as S - F = (1/2)*M*a. The tension and friction exert opposing torques. (You'd only write the equation the way you did if you assumed that friction acted to the left. No big deal -- you'll just end up with a negative value for F. Pick a convention for F and stick to it in all your equations.)

11. Apr 1, 2005

### xanthym

Concerning the Friction Force "Ffric" used in the MSG #7 solution, remember that (+) direction is to the right, that "Ffric" was chosen to point left, and that "Ffric" occurs in TWO (2) basic equations:

a) Disk LINEAR Motion (of Disk Center of Mass):
--> M*a = S - Ffric <--- "Ffric" points left, opposes "S" for linear acc of disk center of mass

b) Disk ROTATIONAL Motion (about Rotation Axis)
--> {Net Torque} = Iα = S*R + Ffric*R <--- "Ffric" points left, adds to "S" clockwise torque

"Ffric" was ARBITRARILY chosen to point LEFT, thereby OPPOSING the String's Tension "S" for the Disk's LINEAR acceleration (of Disk Center of Mass), but ADDING TO the String Tension TORQUE "R*S" for the Disk's ROTATIONAL (clockwise) acceleration. It's use is consistent in the above equations. It can be chosen either way, provided it's applied consistently.

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