How much force would a 55kg person experience on a roller coaster if they travel through a loop with a radius of 55kg at a speed
of 14.1 m/s?
1 answer:
Answer:
200 N
Explanation:
For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.
The radially inward acceleration, or the centripetal acceleration is given by :
a = v² / r
where v is the speed at which the body is moving and r is the radius of the circle
Given-
m = 55kg
v = 14.1 m/s
r= 55m
We know that F = ma
⇒ F = m ( v²/ r )
⇒ F = 55 x 14.1 x 14.1 / 55
⇒ F =14.1 x 14.1 = 200 N
∴ <em>The force acting is 200 N</em>.
You might be interested in
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
