2019 A/L Physics Past Paper · Question 50 (English Medium)

An object inside a rotating cone — the maximum angular velocity

A small object is placed inside a right circular cone with axis vertical and vertex down as shown in the figure. The coefficient of static friction between the inner surface of the cone and the object is μ\mu. What is the maximum angular velocity of rotation of the cone about its axis for the object to be on the inner surface of the cone without slipping at a distance dd away from the axis?

  1. (1) g(cos⁡θ−μsin⁡θ)d(sin⁡θ+μcos⁡θ)\sqrt{\frac{g(\cos\theta - \mu\sin\theta)}{d(\sin\theta + \mu\cos\theta)}}
  2. (2) g(sin⁡θ−μcos⁡θ)d(cos⁡θ+μsin⁡θ)\sqrt{\frac{g(\sin\theta - \mu\cos\theta)}{d(\cos\theta + \mu\sin\theta)}}
  3. (3) g(cos⁡θ+μsin⁡θ)d(sin⁡θ−μcos⁡θ)\sqrt{\frac{g(\cos\theta + \mu\sin\theta)}{d(\sin\theta - \mu\cos\theta)}}
  4. (4) g(sin⁡θ+μcos⁡θ)d(cos⁡θ−μsin⁡θ)\sqrt{\frac{g(\sin\theta + \mu\cos\theta)}{d(\cos\theta - \mu\sin\theta)}}
  5. (5) gdtan⁡θ\sqrt{\frac{g}{d\tan\theta}}
Answer and explanation

Answer: (3) g(cos⁡θ+μsin⁡θ)d(sin⁡θ−μcos⁡θ)\sqrt{\frac{g(\cos\theta + \mu\sin\theta)}{d(\sin\theta - \mu\cos\theta)}}

What the question asks

We want:

The biggest angular velocity ω\omega at which the object does not slide.

Give it in terms of gg, dd, θ\theta and μ\mu.

Relevant theory

Thus the acceleration of a body performing circular motion is directed towards the centre of the circle and is hence known as "centripetal acceleration".

The frictional force acting when the surfaces just begin to slip past one another is called the limiting frictional force.

The object moves in a circle of radius dd.

So the push on it towards the axis is mω2dm\omega^2 d.

At the biggest ω\omega the object is about to slide up.

So friction acts down the wall. It is μN\mu N.

Required equations

Nsin⁡θ−μNcos⁡θ=mgN\sin\theta - \mu N\cos\theta = mg

Ncos⁡θ+μNsin⁡θ=mω2dN\cos\theta + \mu N\sin\theta = m\omega^2 d

Reasoning and calculation

The letters:

θ\theta: the angle between the wall and the vertical axis.

NN: the push of the wall, in and up.

μN\mu N: the friction, down the wall.

Split NN and μN\mu N into up-down and sideways parts.

NN: up Nsin⁡θN\sin\theta, in Ncos⁡θN\cos\theta.

μN\mu N: down μNcos⁡θ\mu N\cos\theta, in μNsin⁡θ\mu N\sin\theta.

Divide the second equation by the first.

NN and mm go away:

ω2dg=cos⁡θ+μsin⁡θsin⁡θ−μcos⁡θ\dfrac{\omega^2 d}{g} = \dfrac{\cos\theta + \mu\sin\theta}{\sin\theta - \mu\cos\theta}

ωmax=g(cos⁡θ+μsin⁡θ)d(sin⁡θ−μcos⁡θ)\omega_{max} = \sqrt{\frac{g(\cos\theta + \mu\sin\theta)}{d(\sin\theta - \mu\cos\theta)}}

Reading the options

Look at the answers.

The result matches option (3).

Option (1) has friction up the wall. That is the smallest speed.

Option (5) is the case μ=0\mu = 0.

Checks

Check the unit.

gd\dfrac{g}{d} has the unit m s−2m=s−2\dfrac{\text{m s}^{-2}}{\text{m}} = \text{s}^{-2}.

Its root is s−1\text{s}^{-1}, that is rad s−1\text{rad s}^{-1} ✓.

Does it make sense?

If μ=0\mu = 0, we get ω=gdtan⁡θ\omega = \sqrt{\dfrac{g}{d\tan\theta}}. That is option (5) ✓.

If μ\mu gets close to tan⁡θ\tan\theta, the bottom of the fraction goes to zero.

Then the object never slides, at any speed ✓.

Conclusion

ωmax=g(cos⁡θ+μsin⁡θ)d(sin⁡θ−μcos⁡θ)\omega_{max} = \sqrt{\frac{g(\cos\theta + \mu\sin\theta)}{d(\sin\theta - \mu\cos\theta)}}

Friction acts down the wall:

ωmax=g(cos⁡θ+μsin⁡θ)d(sin⁡θ−μcos⁡θ)\omega_{max} = \sqrt{\dfrac{g(\cos\theta + \mu\sin\theta)}{d(\sin\theta - \mu\cos\theta)}}

So the answer is option (3).

Taking friction up the wall even at the biggest ω\omega.

Friction opposes the way the object tends to slide. At a high ω\omega it tends to slide up. So friction acts down.

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