The roller coaster starts with a velocity of 16 m/s. One of the riders is a small girl of mass 26 kg. Find her apparent weight when the roller coaster is at locations B and C. At these two locations, the track is circular, with the radii of curvature given.
rB = 10 m,
rC = 20 m,
The heights at points A, B, and C are
hA = 22 m,
hB = 32 m,
and
hC = 0.
Assume friction is negligible and ignore the kinetic energy of the wheels. (The figure is not necessarily drawn to scale.)
What is the weight at B?
Point A: h₁= 22 m, v₁=16 m/s;
Point B: h₂= 32 m, r₂= 10 m, v₂=?;
Δh₁₂ = h₂-h₁=32-22 = 10 m;
Point C: h₃=0, r₃= 20 m, v₃=?;
Δh₂₃ = h₂-h₃ = 32 – 0 = 32 m.
KE₁+PE₁=PE₂ + KE₂,
mv₁²/2 +mgh₁=mgh₂ +mv₂²/2,
mv₁²/2= mv₂²/2+ mgh₂ - mgh₁=
=mv₂²/2+ mg Δh₁₂.
v₂ = sqrt{ v₁²- 2gΔh₁₂} =
=sqrt{16² - 2•9.8•10} =7.75 m/s.
For point B:
ma₂=mg-N₂,
N₂=mg-ma₂=m[g- (v₂²/r₂)]=
= 26[ 9.8 – (7.75²/10)]= 98.8 N,
N₂ (normal force) = W₂ (weight)
W₂= 98.8 N.
PE₂ +KE₂ = PE₃+KE₃,
mgh₂ +mv₂²/2 = mgh₃ +mv₃²/2,
mv₃²/2 = mgh₂ +mv₂²/2 - mgh₃=
=mv₂²/2 + mg Δh₂₃ ,
v₃ = sqrt{v₂² +2g Δh₂₃}=
=sqrt{ 7.75² + 2•9.8•32) = 26.2 m/s.
For point C:
ma₃= N₃ – mg,
N₃= mg+ma₃=m[g+ (v₃²/r₃)]=
= 26[ 9.8 + (26.2²/20)]= 1148.2 N.
N₃ (normal force) = W₃ (weight)
W₃= 1148.2 N.
To find the apparent weight of the small girl at location B on the roller coaster, we need to consider the normal force acting on her.
At location B, the roller coaster is moving in a circular path with a radius of curvature of 10 m. We know that the girl's mass is 26 kg and her weight is given by the formula:
Weight = mass × gravity
where gravity is approximately 9.8 m/s².
Weight = 26 kg × 9.8 m/s² = 254.8 N
However, since the roller coaster is in circular motion, there is also a centripetal force acting towards the center of the circle. This force is provided by the normal force.
Using the concept of apparent weight, we can relate the normal force (N) to the weight (W) and the centripetal force (Fc):
N = W + Fc
At location B, the centripetal force can be calculated using the formula:
Fc = (mass × velocity²) / radius
where velocity is the speed of the roller coaster.
Given that the velocity at the start is 16 m/s and the radius at B is 10 m, we can calculate the centripetal force:
Fc = (26 kg × 16 m/s²) / 10 m = 41.6 N
Now we can substitute the values of weight and centripetal force into the equation for the normal force:
N = 254.8 N + 41.6 N = 296.4 N
Therefore, the weight of the small girl at location B is 296.4 N.
To find the apparent weight of the small girl at point B, we need to consider the forces acting on her at that location. The apparent weight is essentially the normal force exerted by the roller coaster on the girl.
At point B, the girl is undergoing circular motion, so there must be a centripetal force acting towards the center of the circular path. This centripetal force is provided by the normal force. We can calculate the normal force using the following steps:
1. Determine the net force on the girl at point B.
2. Equate the net force with the centripetal force.
3. Calculate the apparent weight as the magnitude of the normal force.
Step 1: Determine the net force on the girl at point B
The only forces acting on the girl at point B are her weight (mg) and the normal force (N). The net force is the vector sum of these forces and is directed towards the center of the circular path. Since the girl is not accelerating vertically at point B, the net force is zero (N = 0).
Step 2: Equate the net force with the centripetal force
The net force (N) acting on the girl at point B is equal to the centripetal force (F centripetal). The centripetal force is given by:
F centripetal = (mass × velocity^2) / radius of curvature
Given:
- Mass of the girl (m) = 26 kg
- Velocity of the roller coaster (v) = 16 m/s
- Radius of curvature at point B (rB) = 10 m
F centripetal = (26 kg × (16 m/s)^2) / 10 m
= 26 kg × 256 m^2/s^2 / 10 m
= 6656 N
Step 3: Calculate the apparent weight as the magnitude of the normal force
Since the normal force (N) provides the centripetal force, it is equal to the magnitude of the centripetal force. Therefore, the apparent weight of the girl at point B is 6656 N.
So, the apparent weight of the small girl at point B is 6656 N.