How do I find the spinning speed for a block when it is moving in circles?

Chemist116

The problem is as follows:

The figure from below shows a block of mass $m$ and an incline. The frictional force is $20\,N$. Find the rotation speed in $\frac{m}{s}$ necessary such as the block begins to move upwards in the incline.​

The alternatives given in my book are:

$\begin{array}{ll} 1.&\sqrt{5}\,\frac{m}{s}\\ 2.&2\sqrt{5}\,\frac{m}{s}\\ 3.&\frac{3\sqrt{5}}{5}\,\frac{m}{s}\\ 4.&\sqrt{3}\,\frac{m}{s}\\ 5.&2\sqrt{3}\,\frac{m}{s}\\ \end{array}$

I'm stuck with this problem as I don't know how to relate the frictional force such as the block slides up in the incline.

The only formula which comes to my mind in this kind of situation is this:

$v=\sqrt{Rg\tan\omega}$

Provided that masses cancel and a block is down in a incline. But I don't know how can I order the information to make it useful.

Can somebody help me here?. What can I do?. How to solve this problem?. :help:

DarnItJimImAnEngineer

Centripetal acceleration is $\omega r^2$ ($\omega L^2$ in the diagram), acting horizontally towards the centre. The block starts moving when the horizontal component of the frictional force no longer exceeds the centripetal force necessary to hold the block in place.

That said, I don't know what the question is asking for. Angular speed can be given in $\frac{rad}{s}$, $\frac{rev}{s}=Hz$, $\frac{rev}{mn}$, etc., but not in $\frac{m}{s}$.

skeeter

Math Team
The equation, $v = \sqrt{Rg\tan{\alpha}}$ is the speed of an object rotating on banked surface where no friction is required to maintain its position. If there is no friction, $v = \sqrt{L g \tan{\alpha}} = \sqrt{3} \text{ m/s}$

Since there is friction, the block can remain in its same position on the bank at higher speed since friction will contribute to the centripetal force.

Refer to the attached diagram ...

$F_c = \dfrac{mv^2}{L} = N\sin{\alpha} + f\cos{\alpha}$

equilibrium in the vertical direction $\implies N\cos{\alpha} = mg+f\sin{\alpha}$

therefore, $F_c = (mg+f\sin{\alpha}) \cdot \tan{\alpha} + f\cos{\alpha}$

$F_c = \dfrac{mv^2}{L} \implies v = \sqrt{\dfrac{F_c \cdot L}{m}} = \sqrt{5} \text{ m/s}$

Once again, the question posed ...

Find the rotation speed in m/s necessary such as the block begins to move upwards in the incline.
... makes little sense in the context of the given answer choices. The speed required to move up the incline must be greater than $\sqrt{5} \text{ m/s}$. There are two speeds in the given choices that meet that criteria, #2 & #5.

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DarnItJimImAnEngineer

To me, the words "necessary" and "begins" imply we are looking for the lowest speed that satisfied the criterion, i.e., #1.

I see now they were talking about tangential speed of the block. I will try to quiet the little voice in my head yelling about how that's stupid because angular velocity of the assembly is so much easier to measure/control in this instance than linear velocity at the block's position.

1 person

skeeter

Math Team
To me, the words "necessary" and "begins" imply we are looking for the lowest speed that satisfied the criterion, i.e., #1.

I see now they were talking about tangential speed of the block. I will try to quiet the little voice in my head yelling about how that's stupid because angular velocity of the assembly is so much easier to measure/control in this instance than linear velocity at the block's position.
It seems these problems may have been posed in some language other than English, and some detail(s) may have been misinterpreted in translation.

I agree with your view re: angular vs linear speed.

1 person

Chemist116

The equation, $v = \sqrt{Rg\tan{\alpha}}$ is the speed of an object rotating on banked surface where no friction is required to maintain its position. If there is no friction, $v = \sqrt{L g \tan{\alpha}} = \sqrt{3} \text{ m/s}$

Since there is friction, the block can remain in its same position on the bank at higher speed since friction will contribute to the centripetal force.

Refer to the attached diagram ...

$F_c = \dfrac{mv^2}{L} = N\sin{\alpha} + f\cos{\alpha}$

equilibrium in the vertical direction $\implies N\cos{\alpha} = mg+f\sin{\alpha}$

therefore, $F_c = (mg+f\sin{\alpha}) \cdot \tan{\alpha} + f\cos{\alpha}$

$F_c = \dfrac{mv^2}{L} \implies v = \sqrt{\dfrac{F_c \cdot L}{m}} = \sqrt{5} \text{ m/s}$

Once again, the question posed ...

... makes little sense in the context of the given answer choices. The speed required to move up the incline must be greater than $\sqrt{5} \text{ m/s}$. There are two speeds in the given choices that meet that criteria, #2 & #5.
I feel so dumb for not noticing the vectors which you included in the diagram. By comparing to what I did in my notes I spotted that I got incorrect the direction of the frictional force which is pointing downwards. I incorrectly assigned it upwards.

As it was mentioned by:

DarnItJimImAnEngineer said:
To me, the words "necessary" and "begins" imply we are looking for the lowest speed that satisfied the criterion, i.e., #1.
You arrived to the right answer as compared with the answers sheet.

DarnItJimImAnEngineer said:
I see now they were talking about tangential speed of the block. I will try to quiet the little voice in my head yelling about how that's stupid because angular velocity of the assembly is so much easier to measure/control in this instance than linear velocity at the block's position.
As you pointed out (yes) they were indeed talking about the tangential speed. The second comment you made I don't get it. How exactly would you measure the angular speed?. Perhaps do you mean a Tachometer?. I remember having seen similar examples in my electronics cookbook but never attempted making one. About controling the rotational velocity, yes I think this is evident, but again I feel that whoever posed this problem intended to make this as a "fictional" excersise to test conceptual knowledge of how centripetal force can be related with frictional force. This reply also accounts for your first comment which I did also noticed but I felt it didn't applied in this situation, but I like the fact that made me to keep in mind those formulas. :spin:

skeeter said:
It seems these problems may have been posed in some language other than English, and some detail(s) may have been misinterpreted in translation.
I'm so sorry for any source of confusion. As you see, these problems I have to transcribe and translate from the original source in a way that it can be understood, this added to the fact that I do not have much time to review the material and the time I spend in drawing by hand each diagram so it can be seen clearly is very consuming and I have to be stay awake into the wee hours of the morning. But anyways I am very grateful that despite my knowledge often struggles with these concepts once I read your feedbacks makes me realize what I did wrong.

DarnItJimImAnEngineer

Tachometer, yes, but also it's rotating at about 53 rpm (or 0.9 Hz). You could time the rotations by hand.