A 480N sphere 40.0cm in radius rolls without slipping 1200cm down a ramp that is inclined at 53 0 with the horizontal. What is the angular speed of the sphere at the bottom of t…

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asked May 17, 2019 147k views

1 Answer

Drewid · Answer 1 · 2019-05-21T18:17:24+0000

As we know that sphere roll without slipping so there is no loss of energy in this case

so here we can say that total energy is conserved

Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

$(1)/(2)mv_i^2 + mgh = (1)/(2)mv^2 + (1)/(2)I\omega^2+ mgh'$

as we know that ball start from rest

v_i = 0

height of the ball initially is given as

$h = Lsin\theta$

h = 1200sin53 = 960 cm

also we know that

I = (2)/(5)mR^2

also for pure rolling

$v = r\omega$

also we know that

480 = m*9.8

m = 49 kg

now plug in all data in above equation

$480*9.60 + 0 = (1)/(2)*49*(0.40*\omega)^2 + (1)/(2)*(2)/(5)*49*(0.40)^2\omega^2 + 0$

$4608 = 3.92\omega^2 + 1.568\omega^2$

$\omega^2 = 839.65$

$\omega = 29 rad/s$

So speed at the bottom of the inclined plane will be 29 rad/s