Recent content by Mohammad Ishmas

I think the release of 30 degrees is after the vertical such that it makes a acute angle with the horizontal , this is assuming the natural bowling action of any bowler , but physically there can be two cases. And yeah the question is asking horizontal distance only.

oh sorry , I mistyped the formula , it is actually s = rϴ, now we can divide both side by time and then get the relation , linear velocity (v) = rω, here ω is the angular velocity

This problem was given by my physics teacher. I have confirmed that this is the data only , maybe he just gave the data just for the sake of the problem.

Ohh , I got it . We can just divide both sides by time taken and then we'll obtain the relation , thank you .

I think I got you, you are guiding me towards the arc length formula that is , s = r/θ . How does this help with relation between linear velocity and angular velocity ?

I have found an answer in terms of x and y components of the linear velocity of the ball , but I don't know how to relate linear velocity with angular velocity . I know there is a formula regarding it but I want to know the derivation behind it

If this was the question , where the light strikes the slanted surface of the cone then how we would find the area of where the light exerts force ?

Thank you , I understood your point. It would have been more clearer if there were 2 light sources mentioned in the question like one that strikes the top of frustrum and the other strikes the base .

Sorry but I didn't get you, what do you mean by , "the larger surface is partly shaded or displaced by the smaller one." ?

Yeah I did , I used the different formulas for completely reflective and completely absorbent surfaces (for reflective surfaces the force just becomes twice). the final answer comes out to be , Fa/Fb = 2(H/h)^2

I found a solution using this value of 'r', I just plugged this value of "r" in πr^2 , then you would get a ratio of the two areas in terms of height, basically if the area of the base in A and area of top surface is B , then ratio A/B = (H/h)^2 , then we can just use the formulas to find the...

Yeah I can see that , we can use similar triangles to find that r = R(h/H)

The original question is as follows - A uniform beam of light falls on a frustum of a cone. The smaller top surface of the frustum is completely absorbent, while the larger base is perfectly reflective. The original cone, before being cut, had a total height H and a base radius R. The frustum...

Yes this is the question, I am sorry for not giving the diagram as it was not there in the original question

In the original question, radii and heights aren't given. We basically have to just derive a ratio of the forces , and the best approach I could think of was to just find the area of the base in terms of area of the frustrum. It would give us something in terms of either radii or heights , but...