Question

1. a 1200 kg roller coaster car is moving around a vertical circular loop of radius 12.0 m.

a. determine the minimum speed that the roller coaster car needs to move so that it does not fall out of the loop.
b. determine the force of the track acting on the roller coaster car if the car is moving at twice the minimum speed
i. when the car is at the top of the loop.
ii. when the car is at the bottom of the loop.
c. determine the apparent weight of the car when it is at the bottom of the loop at twice the minimum speed.

Explanation:

Part a: Minimum Speed at Top

Step1: Identify Centripetal Force Condition

At the minimum speed, the normal force \( F_N = 0 \), so gravity provides centripetal force: \( F_c = F_g \)
\( \frac{mv^2}{r} = mg \)

Step2: Solve for Speed \( v \)

Cancel \( m \): \( \frac{v^2}{r} = g \)
\( v = \sqrt{gr} \)
Substitute \( g = 9.8 \, \text{m/s}^2 \), \( r = 12.0 \, \text{m} \):
\( v = \sqrt{9.8 \times 12.0} \approx \sqrt{117.6} \approx 10.84 \, \text{m/s} \)

Part b (i): Force at Top (Twice Min Speed)

Step1: Find Speed \( v' = 2v \)

\( v' = 2 \times 10.84 \approx 21.68 \, \text{m/s} \)

Step2: Apply Centripetal Force Formula

At top, \( F_c = F_N + F_g \) (both downward), so \( F_N = \frac{mv'^2}{r} - mg \)
Substitute \( m = 1200 \, \text{kg} \), \( v' = 21.68 \, \text{m/s} \), \( r = 12.0 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \):
\( \frac{mv'^2}{r} = \frac{1200 \times (21.68)^2}{12.0} \approx \frac{1200 \times 470}{12} = 47000 \, \text{N} \)
\( mg = 1200 \times 9.8 = 11760 \, \text{N} \)
\( F_N = 47000 - 11760 = 35240 \, \text{N} \) (or use \( v' = 2\sqrt{gr} \), so \( \frac{mv'^2}{r} = 4mg \), thus \( F_N = 4mg - mg = 3mg = 3 \times 11760 = 35280 \, \text{N} \) (minor difference due to rounding))

Part b (ii): Force at Bottom (Twice Min Speed)

Step1: Centripetal Force Direction

At bottom, \( F_c = F_N - F_g \) ( \( F_N \) upward, \( F_g \) downward), so \( F_N = \frac{mv'^2}{r} + mg \)

Step2: Calculate \( F_N \)

Using \( \frac{mv'^2}{r} = 4mg \) (from \( v' = 2\sqrt{gr} \)):
\( F_N = 4mg + mg = 5mg = 5 \times 11760 = 58800 \, \text{N} \)

Part c: Apparent Weight at Bottom (Twice Min Speed)

Step1: Apparent Weight = Normal Force

At bottom, apparent weight equals the normal force \( F_N \) (from part b ii).

Step2: Substitute Values

\( F_N = 5mg = 5 \times 1200 \times 9.8 = 58800 \, \text{N} \)

Answer:

s:
a. Minimum speed: \( \boldsymbol{\approx 10.8 \, \text{m/s}} \) (exact: \( \sqrt{117.6} \approx 10.84 \, \text{m/s} \))
b (i). Track force at top: \( \boldsymbol{\approx 35300 \, \text{N}} \) (or \( 3mg = 35280 \, \text{N} \))
b (ii). Track force at bottom: \( \boldsymbol{58800 \, \text{N}} \) (or \( 5mg \))
c. Apparent weight at bottom: \( \boldsymbol{58800 \, \text{N}} \) (same as track force at bottom)