We know that
g = LcosΘ
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>
Gravity is also called gravitation
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as


Now the normal force on the block is given as



now the friction force on the cart is given as



now if cart moves with constant speed then net force on cart must be zero
so now we have




so the force must be 199.2 N
Answer:
r = 58.44 [m]
Explanation:
To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.
a = v²/r
where:
a = centripetal acceleration = 15.4 [m/s²]
v = tangential speed = 30 [m/s]
r = radius or distance [m]
r = v²/a
r = 30²/15.4
r = 58.44 [m]