Kinematics In Two Dimensions?

Question

How to solve the belowing question?
A solid sphere with a diameter of 0.17m is released from rest, it then rolls without slipping down a ramp, dropping through a vertical height of 0.61m. The ball leaves the bottom of the ramp, which is 1.22m above the floor, moving horizontally.
Through what horizontal distance(d) does the ball move before landing?
How many revolutions does the ball make during its fall?

Answer 1

not clear what is moving horizontally: floor, ramp or ball. I’m assuming the ball.
♦ at the bottom of the ramp ball’s kinetic energy E=E1+E2,
where E1=0.5*m*v^2 is KE of linear motion, E2=0.5*J*w^2 is KE of rotation, J=0.4·m·r^2 is inertia moment of the ball, w=v/r is angular speed, v is linear speed of the ball, r=0.17/2=0.085m is radius, m is ball’s mass;
♠ on the other hand at the top of the ramp ball’s potential energy was E=mgh, where h=0.61m; as energies conserve E=E, then
mgh= 0.5*m*v^2 +0.5*J*w^2; or;
2gh= v^2 +(0.4*r^2)*(v/r)^2, hence v = √(2gh/1.4);
♣ now the ball is moving at constant horizontal speed v, while its vertical path y=1.22m = 0.5*g*t^2, hence time of free fall t=√(2y/g);
thus ball’s horizontal path x=v*t = √(2gh/1.4) *√(2y/g) =
=√(2hy/1.4) = √(2*0.61*1.22/1.4) =1.031m;
♦ # of revolutions n=w*t/(2pi), where w=v/r is angular speed, and time of free fall t=√(2y/g), and v=√(2gh/1.4); be a good boy to continue; arigato:

Answer 2

a = acceleration along the ramp
g = acceleration of gravity = 9.8 m/s^2
s = length of ramp
h = height of ramp = 0.61 m
t = time to roll down the ramp

First find the acceleration down the ramp. If A is the angle the ramp makes with the horizontal then:
a = g*sin(A) = gh/s since sin(A) = h/s

The time it takes the ball to roll down the ramp is:
s = (1/2)at^2 so t = SQRT(2s/a)

Speed "v" at the end of the ramp is just a*t:
v = a*t = a*SQRT(2s/a) = SQRT(2as) = SQRT(2gh)
v = SQRT(1.22g) = 11.956 m/s

This is independent of the angle of the ramp. Since the problem states that the ball leaves the ramp moving horizontally then this is equal to the horizontal speed. The time it moves is equal to the time it now takes the ball to reach the floor and this is determined by gravity and the height the ramp is above the floor.

s = (1/2)gt^2
1.22 = (1/2)(9.8)t^2
t = SQRT(2.44/9.8) = 0.5 seconds

So the ball moves for 0.5 seconds. Then the horizontal distance is this time times the speed at the bottom of the ramp.

Distance = (0.5)(11.956) = 5.97 meters

The ball is 0.17 in diameter.
Circumference (C) of the ball is pi*d = 0.534 m
Angular speed = w = v/C = 11.956 / 0.534 = 22.386 revs/sec

Number of revolutions = w*t = 22.386*0.5 = 11.19

The ball makes 11.19 revolutions during it's fall to the ground

Answer 3

this is been a on an identical time as, yet listed under are some clues to get began. destroy the preliminary action into its 2 components. You try this via multiplying the speed via the sin (for vertical) and cosine (for horizontal) of the preliminary attitude. Gravity is a persevering with downward rigidity of 10m/sec^2, approximately, so which you comprehend the vertical speed at any time t after the initiating is sixty 5.3*sin(34.5) - 10*t. it fairly is 0 whilst the projectile is on the optimum element, resolve for t. With t you are able to compute the optimum top, and the entire time ( = 2*t, i think for simplicity preliminary top is 0) Horizontal speed does not exchange considering the fact that there is not any horizontal rigidity (ignoring air friction), so the gap travelled is sixty 5.3*cos(34.5)*2*t. speed a million.5 seconds after firing: upload horizontal and vertical vectors: (sixty 5.3*cos(34.5))^2 + (sixty 5.3*sin(34.5) - 10*t)^2 = v^2 at time t.