Question

A mass M of 3.80E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of 5.35 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 34.0°.

Answer 1

N = 26.59 N

Step-by-step explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

`(1)/(2)mv_t^2 + mg(2R) = (1)/(2)mv^2 + mgR(1-cos \theta)`

`mv^2 = mv_t^2 + 2mgR(1 + cos \theta)`

we know that,

`N - mgcos \theta = (mv^2)/(R)`

`N - mgcos \theta = (mv_t^2 + 2mgR(1 + cos \theta))/(R)`

`N - mgcos \theta = (mv_t^2 )/(R)+ 2mg(1 + cos \theta)`

`N  = (mv_t^2 )/(R)+ 2mg + 3mgcos \theta)`

`N  = (0.38* 5.35^2 )/(1.1)+ 2* 0.38* 9.8 + 3* 0.38 * 9.8 cos 34^0)`

N = 26.59 N