Answer:
v = 0.56 m/s
Explanation:
When passing the loop-the-loop, the car is making circular motion.
where R = 33/2 = 16.5 mm = 0.0165 m.
The acceleration can be found by Newton’s Second Law: F = ma.
At the top of the loop-the-loop: F = N + mg = 2mg.
Combining the two equations yields:
[tex]2mg = m\frac{v^2}{R}\\
2g = \frac{v^2}{R}\\
2(9.8) = \frac{v^2}{0.0165}\\
v = 0.56~m/s[/tex]
