A toy car moves around a loop-the-loop track. The loop is
0.5 m high.What is the minimum speed of the car at the top of
the loop for it to stay on the track?
at the top of the loop , gravity is supplying the centripetal force
m * v^2 / r = m * g ... v^2 = r * g
No
Well, if the toy car wants to stay on the track, it better hold on tight and not go flying off into the great unknown. To make sure of that, the car needs to have enough speed at the top of the loop to counteract the gravitational force pulling it downwards. To calculate the minimum speed, we can use a fun equation called the "Critical Velocity Equation."
The equation goes like this: V = sqrt(R * g), where V is the minimum speed required, R is the radius of the loop (in this case, the radius of a vertical loop is equal to the height of the loop), and g is the acceleration due to gravity.
In this case, we have a loop that is 0.5 meters high, so that's the radius. And gravity, well, it's always 9.8 m/s² on Earth.
Plugging in the numbers, we get V = sqrt(0.5 * 9.8). So, let me grab my calculator and do some quick math...
*dramatic calculator typing sounds*
The minimum speed for the car to stay on the track is approximately 3.13 m/s! So, make sure that little car is ready to go vroom vroom at that speed at the top of the loop, or else it might end up taking a detour to outer space!
To find the minimum speed of the car at the top of the loop for it to stay on the track, we can use the principle of conservation of energy.
At the top of the loop, the car has both kinetic energy (due to its motion) and potential energy (due to its height relative to the ground). The minimum speed occurs when the car just barely stays on the track.
At the top of the loop, all of the car's energy is in the form of potential energy. The potential energy at the top of the loop is given by the equation:
Potential energy = mass * gravity * height
The minimum speed can be calculated using the equation:
Potential energy = Kinetic energy
Therefore:
mass * gravity * height = (1/2) * mass * velocity^2
where:
mass = mass of the car
gravity = acceleration due to gravity (approximately 9.8 m/s^2)
height = height of the loop (0.5 m)
velocity = minimum velocity at the top of the loop
Simplifying the equation:
velocity^2 = 2 * gravity * height
Substituting the values:
velocity^2 = 2 * 9.8 m/s^2 * 0.5 m
velocity^2 = 9.8 m^2/s^2
Taking the square root of both sides:
velocity = √(9.8 m^2/s^2)
velocity ≈ 3.13 m/s
Therefore, the minimum speed of the car at the top of the loop for it to stay on the track is approximately 3.13 m/s.