Colliding Rod!

Classical Mechanics Level 5

A uniform rod of length $L = 3\text{ m}$ and Mass $M = 3.1565$ is aligned with horizontal direction at an angle $\theta = 37^\circ$ at a height of $H = 9.35\text{ m}$ from horizontal. Now it is Released from that position . It collides with the sufficiently rough floor such that just after collision it's vertical velocity reduces to zero. Find the angular velocity ( $\omega$ ) of the Rod just after collision .

Details And Assumptions :

The ground is sufficiently rough to prevent slipping of contact point.
Take $g = 10\text{ m/s}^2$ .
Initial height given is of centre of mass of rod from ground .

Hint : Get answer in variables first and then put data .

Original

This is a part of my set Aniket's Level 5 Challenges in Classical Mechanics

The answer is 10.0.

2 solutions

Aniket Sanghi
Aug 10, 2016

A better way would be to conserve angular momentum about the ground points on which impact occurs.

Rajdeep Dhingra - 4 years, 10 months ago

@Rajdeep Dhingra once you give a brief about your method also! (Edit sol)

I liked your method! 👍

Aniket Sanghi - 4 years, 7 months ago

But in question it is said no slipping

Ishan Dixit - 4 years, 2 months ago

@Ishan Dixit – Because of Friction only there is no slipping . The net velocity of contact point is zero. Horizontal velocity neednt be zero.

Aniket Sanghi - 4 years, 2 months ago

@Aniket Sanghi – Thanks got it

Ishan Dixit - 4 years, 2 months ago

Finally solved my first level 5 CM question on brilliant! Nice question!

Harsh Shrivastava - 4 years, 7 months ago

That's nice 👍

Aniket Sanghi - 4 years, 7 months ago

I couldn't understand why rod after collision has horizontal velocity plz explain

Ishan Dixit - 4 years, 2 months ago

Because friction provides a horizontal impulse.

Harsh Shrivastava - 4 years, 2 months ago

But in question it is said no slipping

Ishan Dixit - 4 years, 2 months ago

Hats off to the data that you have set @Aniket Sanghi ..BRILLIANT

Ankit Kumar Jain - 3 years, 3 months ago

Prakhar Bindal
Aug 11, 2016

Although Aniket's solution is absolutely correct let me elaborate a bit

Firstly Observe when the rod falls from a certain height there is no torque about the centre of mass hence the rod won't rotate and will do pure translation motion under gravity and strike the ground with some velocity . Be careful here the distance covered by rod's centre is not H It is Slightly less than it which can be found out using elementary geometry . the values provided are too good that weird decimals cancel out and you get integer velocity = 13 on striking.

Now Focus ground will provide impulsive normal reaction as a result of which the frictional force will also be impulsive. in front of these impulsive forces we can neglect all other finite forces like gravity acting on the rod.

Let the rod's centre of mass a horizontal velocity v and rod rotates about it centre of mass with an angular velocity w .

Using Impulse Momentum theorem we have

Ndt = 13m

fdt = mv

where Ndt and fdt denote the normal reaction and frictional impulse acting on the rod.

Then we use Angular equivalent of Impulse Momentum theorem about centre of rod

Ndt*(l/2)cos37 + fdt(l/2)cos53 = Kw

where K Denotes Moment of inertia of rod about its centre and l denotes length of rod

Finally we need one more equation to evaluate w and that has been given that friction is sufficent so that it can prevent slipping at the point of contact

therefore l/2(w)cos(53)+v =0

Solving these and putting values you obtain 10 rad/sec

One suggestion bro please write friction is sufficient to prevent slipping instead of sufficient friction because technically speaking sufficient has relative meanings which might confuse some people (including me)

By the way a nice revision of Rotational mechanics!

A better way would be to conserve angular momentum about the ground points on which impact occurs.

Rajdeep Dhingra - 4 years, 10 months ago

Can we just conserve L about the point of collision? I think we can, since the angular impulse provided by gravity would be negligible as compared to the one provided by contact force, so L would be conserved.

Kushal Thaman - 1 year, 10 months ago

Colliding Rod!

This is a part of my set Aniket's Level 5 Challenges in Classical Mechanics

The answer is 10.0.

2 solutions

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