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?
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.
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...
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.
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.]
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...
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.
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...
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...
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...
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...
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.
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...
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 ##