Solving a Physics Problem Involving a Hinged Stick and Clay

In summary, the conversation discusses a problem with a question involving a thin stick and a piece of clay colliding. The angular velocity of the stick after the collision is found to be 6.79×10-1 rad/s. The ratio of final mechanical energy to initial mechanical energy is incorrectly calculated, which is resolved by using conservation of angular momentum. The conversation also clarifies the difference between moment of inertia about the center of mass and about a different axis, and the use of parallel axes theorem in calculating moment of inertia.
  • #1
zhenyazh
56
0
Hi,
i am preparing for the test and have a problem with the following question.
an image is attached

A thin stick of mass M = 4.9 kg and length L = 2.6 m is hinged at the top. A piece of clay, mass m = 1.0 kg and velocity V = 4.9 m/s hits the stick a distance x = 2.20 m from the hinge and sticks to it. What is the angular velocity of the stick immediately after the collision?

this i found 6.79×10-1 rad/s

What is the ratio of the final mechanical energy to the initial mechanical energy?

if i understand correctly, the final mechanical energy is before the bar starts to move up.
so i should take into consideration only the kinetic energy.
the initial energy is of course the energy of the clay which i regard as a point body
and thus 0.5mv^2.
the final energy is the energy of bar and clay joined.
this is 0.5Iw^2+0.5mu^2 where u is the velocity of the center of mass of the joint body.
if i know that angular speed of the joint body i can use w=ur, where r is the distance
between the centre of mass and the point with respect to which i am doing my calculations. that would be the distance from the pivot that holds the bar.
Xcm=(mx+lM)/(m+M)=2.532
that gives me
0.5Iw^2+0.5mw^2*2.532^2.
but i get the wrong answer.
where am i wrong?

thanks
 

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  • #2
zhenyazh said:
if i understand correctly, the final mechanical energy is before the bar starts to move up.
so i should take into consideration only the kinetic energy.
the initial energy is of course the energy of the clay which i regard as a point body
and thus 0.5mv^2.
the final energy is the energy of bar and clay joined.
Correct so far.
this is 0.5Iw^2+0.5mu^2 where u is the velocity of the center of mass of the joint body.
Not correct. The energy after the collision is just (1/2)Iω2 where I is the moment of inertia of the composite body about the hinge. Use conservation of angular momentum about the hinge to relate the initial velocity of the clay to ω.

This is an inelastic collision that does not conserve mechanical energy.
 
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  • #3
i will try. but can u explain when the energy is a sum of two expression and when not.
that is why do we lose the second part here
 
  • #4
well it worked! :)
so thanks.
but why is it only what u said?
 
  • #5
Consider a rigid body that rotates about a fixed axis. Think of the body consisting of N point masses. All the masses have the same angular velocity ω. The total kinetic energy of these masses is

[tex]K=\sum^{N}_{i=1}\frac{1}{2}m_i v^{2}_{i}[/tex]

But

[tex]v_{i}=\omega r_i[/tex]

Therefore

[tex]K=\sum^{N}_{i=1}\frac{1}{2}m_i \omega^2 r^{2}_{i}=\frac{1}{2} \left( \sum^{N}_{i=1}m_i r^{2}_{i} \right) \omega^2=\frac{1}{2}I \omega^2[/tex]

That's why.
 
  • #6
great thanks. this seems logical.
but i saw in one of my books that the overall kinetic energy
of a body is the kinetic energy of its center of mass that is 0.5mv^2
plus the component with inertia that u mentioned.

why the difference, is there some nuance i am missing?
 
  • #7
Yes. You are missing that there are two ways to write the kinetic energy of a body that instantaneously rotates and translates

Way 1
[tex]K=\frac{1}{2}I \omega^2[/tex]

where I is the moment of inertia about the instantaneous axis of rotation

Way 2
[tex]K=\frac{1}{2}I_{CM} \omega^2+\frac{1}{2}m V_{CM}^2[/tex]

where ICM is the moment of inertia about the center of mass and VCM is the translational speed of the center of mass.

You can derive Way 2 from Way 1 by using the parallel axes theorem. These two are equivalent but you have to pay attention which one you use and what you use for moment of inertia. You cannot mix and much as you seem to have done when you originally attempted this problem.
 
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  • #8
so for example for a rod with a pivot in the middle
would 1/12ml^2 be the I or the Icm?
 
  • #9
Yes, and because it is pivoted in the middle, i.e. at the center of mass, VCM = 0 and the expressions in Way 1 and Way 2 become identical.
 
  • #10
i think u misunderstood me.
what is the difference between the general inetria of a rod and Icm?
 
  • #11
oh and of course really thanks for helping for such long time
 
  • #12
OK, I see the question now, I did misunderstand you. ICM is the moment of inertia about an axis that goes through the center of mass; I is the moment of inertia about any other axis that does not go through the center of mass.

For a rod of mass m and total length L the CM is at L/2 and

ICM=(1/12)mL2

If you want to calculate the moment of inertia about one end of the rod, you have to use the parallel axes theorem. The result is

IEnd=ICM+m(L/2)2=(1/3)mL2.
 
  • #13

1. How do you calculate the net torque on a hinged stick?

The net torque on a hinged stick can be calculated by summing up the torques of all the external forces acting on the stick. The formula for torque is torque = force x distance from the pivot point. So, you would need to identify all the forces acting on the stick, determine their distances from the pivot point, and then calculate their individual torques. Finally, add up all the torques to get the net torque.

2. What is the equilibrium condition for a hinged stick with clay attached?

The equilibrium condition for a hinged stick with clay attached is that the net torque and net force acting on the system must both be equal to zero. This means that the stick is not rotating and is also not moving in any direction. Mathematically, this can be represented as ΣF = 0 and Στ = 0, where Σ indicates the sum of all forces or torques.

3. How does the mass of the clay affect the equilibrium of the hinged stick?

The mass of the clay attached to the hinged stick affects the equilibrium by changing the distribution of weight and the distance from the pivot point. This, in turn, changes the torque exerted by the clay on the stick. A heavier clay will exert a larger torque, which needs to be balanced by an equal and opposite torque to maintain equilibrium.

4. Can the hinged stick and clay system be in both rotational and translational equilibrium at the same time?

Yes, the hinged stick and clay system can be in both rotational and translational equilibrium at the same time. This means that the system is not rotating and is also not moving in any direction. However, it is important to note that the equilibrium equations for rotational and translational motion are different and need to be satisfied separately.

5. How does the angle of the stick affect the torque calculations in this problem?

The angle of the stick affects the torque calculations by changing the distance between the pivot point and the line of action of the force. The torque exerted by a force is directly proportional to this distance. So, if the angle of the stick changes, the torque calculations will also change accordingly.

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