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

Answer:

N = 26.59 N

Explanation:

given,

mass = 0.38 kg

radius of the hoop = 1.10 m

speed = 5.35 m/s

force = ?

now,

$\dfrac{1}{2}mv_t^2 + mg(2R) = \dfrac{1}{2}mv^2 + mgR(1-cos \theta)$

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

we know that,

$N - mgcos \theta = \dfrac{mv^2}{R}$

$N - mgcos \theta = \dfrac{mv_t^2 + 2mgR(1 + cos \theta)}{R}$

$N - mgcos \theta = \dfrac{mv_t^2 }{R}+ 2mg(1 + cos \theta)$

$N  = \dfrac{mv_t^2 }{R}+ 2mg + 3mgcos \theta)$

$N  = \dfrac{0.38\times 5.35^2 }{1.1}+ 2\times 0.38\times 9.8 + 3\times 0.38 \times 9.8 cos 34^0)$

N = 26.59 N