# PhysicsPhysics 12u - Two masses Spinning on a disc

#### Wild ownz al

##### Member
A penny of mass 3.10 g rests on a small 20.0-g block supported by a spinning disk. The block is sitting at the edge of the disc at a radius of 12 cm. If the coefficient of friction between block and disk are 0.750 (static) and 0.640 (kinetic) while those for the penny and
block are 0.450 (kinetic) and 0.520 (static), what is the maximum rate of rotation (in revolutions per minute) that the disk can have before either the block or the penny starts to slip?

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
A penny of mass 3.10 g rests on a small 20.0-g block supported by a spinning disk. If the coefficient of friction between block and disk are 0.750 (static) and 0.640 (kinetic) while those for the penny and
block are 0.450 (kinetic) and 0.520 (static), what is the maximum rate of rotation (in revolutions per minute) that the disk can have before either the block or the penny starts to slip?
Hi wild one,

That depends on the distance of the block and the cent from the axis of rotation.
Is that given?

Either way, since we're talking about 'starts to slip', we're only interested in the static friction.
Since the block has the lower coefficient of static friction, the block will be the first to slip.
The block will start to slip when the centrifugal force on the block is equal to its static friction.

#### Wild ownz al

##### Member
Hi wild one,

That depends on the distance of the block and the cent from the axis of rotation.
Is that given?

Either way, since we're talking about 'starts to slip', we're only interested in the static friction.
Since the block has the lower coefficient of static friction, the block will be the first to slip.
The block will start to slip when the centrifugal force on the block is equal to its static friction.
Hey sorry I missed that. I updated the question including the radius.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey sorry I missed that. I updated the question including the radius.
Hmm... we still need the location of the block...
Is it at the edge of the disk? Then we need to know the size of the block. Is it given?
Or can we assume the disk is a little bigger and the center of the block is at the given radius?

Anyway, let's assume that the block is fully on the disk, and that the center of the block is at a distance $r$ from the axis of rotation.

Then the static friction $F_s$ on the block, just before it starts sliding, is
$$F_{s} = \mu_{s} N\tag 1$$
where $\mu_{s}$ is the coefficient of static friction of the block with the disk, and $N$ is the combined weight of the block and the cent.

The corresponding centripetal force $F_c$ is
$$F_c=m \omega^2 r \tag 2$$
where $m$ is the combined mass of the block and the cent, and $\omega$ is the angular velocity.

The combined weight is
$$N= m g\tag 3$$
where $m$ is again the combined mass, and $g=9.81\,\text{m/s}^2$.

The angular velocity is
$$\omega = 2\pi f\tag 4$$
where $f$ is the frequency, which is the number of revolutions per second.

Finally, we have that
$$f = \frac{rpm}{60} \tag 5$$
where $rpm$ is the revolutions per minute.

Set $F_{s}=F_c$ and solve for $rpm$?

#### Wild ownz al

##### Member
Hmm... we still need the location of the block...
Is it at the edge of the disk? Then we need to know the size of the block. Is it given?
Or can we assume the disk is a little bigger and the center of the block is at the given radius?

Anyway, let's assume that the block is fully on the disk, and that the center of the block is at a distance $r$ from the axis of rotation.

Then the static friction $F_s$ on the block, just before it starts sliding, is
$$F_{s} = \mu_{s} N\tag 1$$
where $\mu_{s}$ is the coefficient of static friction of the block with the disk, and $N$ is the combined weight of the block and the cent.

The corresponding centripetal force $F_c$ is
$$F_c=m \omega^2 r \tag 2$$
where $m$ is the combined mass of the block and the cent, and $\omega$ is the angular velocity.

The combined weight is
$$N= m g\tag 3$$
where $m$ is again the combined mass, and $g=9.81\,\text{m/s}^2$.

The angular velocity is
$$\omega = 2\pi f\tag 4$$
where $f$ is the frequency, which is the number of revolutions per second.

Finally, we have that
$$f = \frac{rpm}{60} \tag 5$$
where $rpm$ is the revolutions per minute.

Set $F_{s}=F_c$ and solve for $rpm$?
Updated the question again with the location of the masses. However there are no sizes of the masses given

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Updated the question again with the location of the masses. However there are no sizes of the masses given
Can it be that we should read it as: "The block is sitting at the edge of the disc at a radius of 12 cm?"

If so, then we can proceed with $r=12\,\text{cm}$.

#### Wild ownz al

##### Member
Can it be that we should read it as: "The block is sitting at the edge of the disc at a radius of 12 cm?"

If so, then we can proceed with $r=12\,\text{cm}$.
That is correct.

#### skeeter

##### Well-known member
MHB Math Helper
Updated the question again with the location of the masses. However there are no sizes of the masses given
your original post gives mass measurements for the penny & block ...

A penny of mass 3.10 g rests on a small 20.0-g block supported by a spinning disk.

#### Wild ownz al

##### Member
your original post gives mass measurements for the penny & block ...
Hey Skeeter. I read his question as if he was looking for the actual dimensions of the objects. The masses of the block and penny are given.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yeah, I meant the actual sizes.
Either way, it does say that the block is 'small', suggesting that we can ignore its size.

So?
How far do you get if you don't mind me asking?