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If ##\frac{d\theta}{dt} = 0## and ##\omega## is nonzero, will the piston move? What do you think? Drawing diagrams may help.

Why didn't you consider ##\omega## in your solution?

You forgot to upload the figure.

Yes, you're correct.

If you know how to set mirror charges for grounded conducting plane and sphere, this picture is enough hint for you.

I guess the boundary is like this in 3D, right? : Do you know how to set up image charge for grounded conducting plane and for grounded conducting sphere?

Calculate the net force in terms of mass and coefficient of friction.

Yes, it was straight forward. I thought the figure might contain some important information. Well, now tell me what is the net force on the object.

You forgot to upload the figure.

Potential energy, $$U_{total}= m_{total} \cdot g\cdot h_{center~of~mass}$$ Or you can calculate the height for the two masses separately, then calculate their respective potential energy and add them.

You misinterpreted the question. The ball moves "within" the slopes plane, I mean, it does not go "above" the plane.

Exactly!

Oh, sorry! I missed the line. If the passenger hold on to something, he does not need to have any velocity at the top of the loop to not fall out. At the ground level, he just needs to have kinetic energy = mass * g * height of the top of the loop The problem we have solved earlier is for the...

Exactly! :thumbup:

The person is pushing the surface (in the direction perpendicular to the surface). That is why the surface is exerting equal normal reaction force on the person in the opposite direction.

You have forgot to consider the potential energy term.

According to the rules of the forum, you have to show your attempt at a solution.

The velocity you have calculated is the minimum required speed 'at the top of the loop'. Are the speeds at the ground level and the top level equal? [Hints: Use energy conservation.]

I think the maximum tension in B is the tension in A minus ##mg##.

Will the mass be accelarated? Won't the rope break if the force exceeds the limit of 50N?

I am confused about how the force will be transmitted.

Won't rope B break?

Then I guess the mass is negligible.

The mass was not given in the problem.

Homework Statement [/B] A body is connected at the middle with two different ropes as in the figure. The maximum force that rope A can resist is 60 N, and for rope B its 50 N. What will happen if someone apply a downward force at the loose end of rope B, a) gradually b) very fast Homework...

Oh, sorry. I misunderstood it.

Thanks for your suggestion. This thought was really interesting. But, I think the pursuit curve problem does not deal with "minimum time" to catch the pursuee.

Oh, yeah, it does. But is there any simple logic behind this fact?

Thanks for your suggestion. But I would like to understand the logic behind the fact that the time would be optimal for the straight path.

I solved in that way and the result matches. But I cannot quite convince myself that the resultant velocity should be in the direction of AC for minimum time. My confusion is that the magnitude of velocity can be higher for other trajectory than this shortest (in terms of length) path.

Thanks for your reply. I took the x-axis in the direction of AB, and the y-axis is perpendicular to it. I calculated the time, ##t = \int_0^{x_1} \frac{1+y'^2}{v_r y' + \sqrt{v_b^2 - v_r^2 + v_b^2 y'^2}} \,dx## [where, ##x_1 = 8~ miles##, ##v_r = speed~ of~ river = 4 ~mi/hr## and ##v_b =...

Homework Statement Two towns A and B, are situated directly opposite to each other on the banks of a river whose width is 8 miles and which flows at speed of 4 mi/hr. A man located at A wishes to reach town C which is 6 miles upstream from and on the same side of the river as town B. If his...

I have got the answer. :)

Homework Statement Light falls on the surface AB of a rectangular slab from air. Determine the smallest refractive index n that the material of the slab can have so that all incident light emerges from the opposite face CD. Homework Equations The Attempt at a Solution Let's think about this...

Here is the question:

Homework Statement In Ruchhardt’s method of measuring ##\gamma##, illustrated in Fig. 12.2, a ball of mass ##m## is placed snugly inside a tube (cross-sectional area ##A##) connected to a container of gas (volume ##V##). The pressure ##p## of the gas inside the container is slightly greater...

Homework Statement Suppose, the following equation describes the relation between an independent and a dependent variable physical quantities(that will be measured by experiments; for example, temperature, current, voltage etc) x & y : ##y = ax^2 + bx + c## We have to find the values of the...

I solved this problem using this technique. Thanks for your help.

Oh! Sorry! How could I make such a silly mistake!:)) Thanks for your help.

x is the expansion of the spring.

Homework Statement Four weightless rods of length ##l## each are connected by hinged joints and form a rhomb (Fig. 48). A hinge A is fixed, and a load is suspended to a hinge C. Hinges D and B are connected by a weightless spring of length ##1.5l## in the undeformed state. In equilibrium, the...

No spatial variation.

Homework Statement Suppose, light is passing through a liquid whose refractive index is time-varying. What will be the path of light ray ? Will it be a straaight line or curve ? Homework Equations The Attempt at a Solution I think, it will be a straight line.

So, it is, I think: for ##m_1## : ##F-T = m_1 a_1## for ##m_2## : ##T = m_2 a_2## the tension on the spring is ##T## ;

I thought, the force will be applied on the whole system, then , ##a = F/(m_1 + m_2)##

Homework Statement Suppose the surface is completely frictionless. Will the spring experience any length change?[/B] Homework Equations The Attempt at a Solution To change the length of the spring, force should be applied from both ends. In this case, there is no force of friction. So, my...

I think, after collision, the rod will have two type of motion: translational and rotational. Let, the final translational velocity of the rod be ##v_1## and final velocity of ball D be ##v_2## ; and final angular velocity of rod be ##\omega_1## ; So, applying the law of (linear) momentum...

That's good! Now, what would happen if the center was not fixed?

In the actual problem, the picture was in the vertical plane. The ball D was falling down with the accelaration of ##9.8 ms^{-2}##. So, in this case, I think, there will be a torque because of collision and another torque which is equal to ##MgR##. Is it so?

So, I think the equation will be, ## 2m \omega r^2 + Mvr = 2m \omega _f r^2 + Mv_f r## And the equation for energy conservation, ##2\cdot \frac {1}{2} m r^2 \omega ^2+ \frac{1}{2} Mv^2 = 2\cdot \frac {1}{2} m r^2\omega _f ^2 + \frac{1}{2} Mv_f ^2 ##