Question

A 1.20 m long ramp is inclined at 20 degree to the horizontal. A solid ball is released from rest at the top of the ramp.

Find the ball's speed at the bottom of the ramp.
Find its translational acceleration.

Answer 1

i tried 1.20*sin20=0.410

then 2(9.8)*(.410)=v^2

=2.83 m/s but this is not right what am i doing wrong for the first part?

i need help for the first n second part please help

Answer 2

gravity is altered by equating g with the the sin

sin(t) = g/a

a = g csc(t) should be a modified acceleration due to gravity

Answer 3

now if i use t for time instead of theta ...

a(t) = -9.8 csc(20)

v(t) = -9.8 csc(20) t

s(t) = -4.9 csc(20) + Ho

since the ramp is h tall and 1.2 long; that gives us Ho = 1.2 sin(20) as a starting point

Answer 4

s(t) = -4.9 csc(20) t^2+ 1.2sin(20)

this reaches the bottom when

t^2 = 1.2sin(20)/4.9csc(20)

Answer 5

with any luck, thats when t = .1892

Answer 6

v(.1892) = -9.8csc(20)*(.1892) = abt -5.422 which is speed in a downward direction if i did it right

Answer 7

i got no idea what a translational accel is tho

Answer 8

unless that is what i determined to begin with, but without knowing the definition its up for grabs

Answer 9

so whys g negative?

Answer 10

* why is

Answer 11

g is negative so that it pulls the object down the ramp

Answer 12

its just shows direction of force

Answer 13

so would that mean that my answer is a negative # as well?

Answer 14

dunno, im pretty much taking a blind stab at this. if i were to take guess at what they want, i would say that they want the positive value to indicate a magnitude only.

Answer 15

i 'd say this question involves rotational dynamics, that is concepts of rolling. (also since its given that the body is a sphere and a solid sphere at that). so, i believe we have to assume that the sphere performs pure rolling and then solve.

Answer 16

i was wondering in the back of my mind if rolling and "frictionless" where similar