Question

1. The problem statement, all variables and given/known data
Consider a frictionless sphere of radius r=5.0m that is fixed to the ground and a small block of mass m=1.0 kg that is nudged slightly from the very top. This takes place on the surface of the Earth. Find the location on the ground where the block lands by following the steps below.

Answer 1

a. Find the numerical value of the angle at which the block leaves the sphere.

b. What is the velocity of the block at the instant it leaves the sphere? (Note the components)

c. Find how far from the sphere's point of contact with the ground (0 on the diagram) the block hits the ground.

Answer 2

This is a pretty interesting problem, but I don't have much time. I can make a start though:
Let's work in a coordinate frame where \(y\) is vertical and \(x\) is horizontal. The ground is the line \(y=0\) and the line \(x=0\) passes through the centre of the sphere
Equation of the circle (sphere in 2D, since the problem is symmetric in the azimuthal angle).
\(x^2+(y-5)^2=25\)
The implicit derivative is
\[\frac{dy}{dx}=\frac{x}{5-y}\]
might be good enough for now, I'm not sure.
The trajectory after leaving the sphere will be
\(y = -a(x-b)^2+c\)
the point at which the block leaves the sphere is \(x=b, y=c\)
The derivative of this is:
\[\frac{dy}{dx}=2a(b-x)\]
This must equal the derivative above of the equation for a circle, since the trajectory must make a smooth transition from moving along the surface to moving along a parabola. Notice that the derivative of \(y\) with respect to \(x\) is the ratio of the velocities. So:
\[\frac{\dot{y}}{\dot{x}} = 2a(b-x) = \frac{x}{5-y}\]
The trajectory in the vertical direction will be
\(y = c + \dot{y} t - 4.9t^2\)
likewise, horizontal trajectory
\(x = b + \dot{x} t\)
We can equate kinetic energy at the point the block leaves the sphere with gravitational potential energy
\(9.8(10-c)=0.5(\dot{x}^2+\dot{y}^2)\)

I think that's all that can be done to set up the problem. Hope it helps somewhat.

Answer 3

hmm that's an interesting way to put it. You took the more mathematical approach, My second approach, which I hope is better was to define the angle as cos(theta)=r-h/r where r is radius and h is the unknown height then work it from there

Answer 4

Note:
Block is in contact with the sphere ⇒ Normal reaction ≠ 0

So, at the point where the block leaves the sphere, perform force balance.
Remember, that the velocity will only have a component tangential to the sphere (why?)

Next perform energy balance between this point and the top most point.

You will get two equations and two unknown variables (namely, v and θ). Solve them.


Now resolve the velocity vector into horizontal and vertical components.
Calculate the time taken for vertical motion and use it two find the distance traveled in horizontal motion.

By simple geometry, find the (horizontal) distance between the lowest point of the sphere and the point at which the block leaves the sphere.