The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 5.0 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times?
Well, if we're talking about loop-the-loops, we better buckle up for some wild physics fun! Now, to keep the car in contact with the track, we need to find the maximum radius r.
Here's the deal: Gravity is a sneaky force, always trying to bring things down. In this case, it's working against the car as it goes through the loop. To counteract gravity, the car needs a minimum speed at the top of the loop.
Now, to find that minimum speed, we can use a little physics equation called centripetal force. This force is what keeps the car moving in a circle.
The centripetal force is given by the equation:
Fc = mv^2 / r
Where Fc is the centripetal force, m is the mass of the car, v is its velocity, and r is the radius of the loop.
We can set this force equal to the force of gravity to find our minimum speed. The force of gravity is given by:
Fg = mg
Where m is the mass of the car and g is the acceleration due to gravity.
Now, here's where the fun begins. The speed at the top of the loop is equal to the initial speed, and we can set the centripetal force equal to the force of gravity:
m(v^2 / r) = mg
The masses cancel out, and we're left with:
v^2 / r = g
Now, we can plug in the numbers. If the initial speed is 5.0 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2, we can rearrange the equation to solve for the maximum radius:
r = v^2 / g
And when we plug in the numbers, we get:
r = (5.0 m/s)^2 / 9.8 m/s^2
r ≈ 2.55 meters
So, the largest value that the radius r can have for the car to stay in contact with the circular track is approximately 2.55 meters. Just remember, no clowning around in physics! Stay safe on those loop-the-loops!
To determine the maximum value for the radius r in order for the car to remain in contact with the circular track throughout the loop-the-loop trick, we need to consider the forces acting on the car at the topmost point of the loop.
1. At the topmost point of the loop, the car's acceleration must be equal to the centripetal acceleration required to keep it in circular motion.
2. The centripetal acceleration (ac) can be calculated using the formula: ac = v^2 / r, where v is the velocity of the car and r is the radius of the circular track.
3. Given that the initial speed of the car is 5.0 m/s, we can substitute this value into the formula: ac = (5.0)^2 / r^2.
4. At the topmost point of the loop, the net force acting on the car is the sum of the gravitational force (mg) and the normal force (N). The net force is directed towards the center of the circular track and is equal to the centripetal force (Fc).
5. The centripetal force can be calculated using the formula: Fc = m * ac, where m is the mass of the car.
6. Since the net force is the sum of the gravitational force and the normal force, we can write the equation: Fc = mg + N.
7. At the topmost point of the loop, the normal force is equal to zero since the track is no longer supporting the car. Therefore, the equation becomes: Fc = mg.
8. Substituting the expression for the centripetal force (Fc) and the mass (m) of the car into the equation: m * ac = mg.
9. Canceling the mass (m) on both sides of the equation: ac = g.
10. Substituting the expression for the centripetal acceleration (ac) into the equation: v^2 / r = g.
11. Rearranging the equation to solve for the maximum value of r: r = v^2 / g.
12. Substituting the given values: r = (5.0)^2 / 9.8.
13. Calculating the value of r: r ≈ 2.55 m.
Therefore, the largest value that the radius r can have for the car to remain in contact with the circular track at all times is approximately 2.55 meters.
To determine the largest value that the radius (r) can have so that the car remains in contact with the circular track, we need to consider the forces acting on the car at the highest point of the loop-the-loop.
The forces acting on the car are the gravitational force (mg) and the centripetal force (mv²/r), where m is the mass of the car, g is the acceleration due to gravity, and v is the velocity of the car. At the highest point of the loop-the-loop, these two forces must add up to provide the required centripetal force.
At the highest point, the gravitational force is pointing downwards and the centripetal force is pointing towards the center of the circle. Therefore, we can add the magnitudes of these forces to find the net force acting on the car.
The gravitational force is given by mg, where m is the mass of the car and g is the acceleration due to gravity (9.8 m/s²).
The centripetal force is given by mv²/r, where m is the mass of the car, v is the velocity of the car (5.0 m/s), and r is the radius of the circular track.
Setting the net force equal to the centripetal force, we have:
mg + mv²/r = mv²/r
Simplifying the equation, we cancel out the mv²/r terms:
mg = 0
This equation implies that the net force is zero, meaning that the car just balances at the top of the loop-the-loop without any force pushing it towards the center of the circle.
Since there is no net force acting on the car at the highest point, the car will remain in contact with the circular track as long as the force of gravity is equal to or greater than the centripetal force. This occurs when the car is at the verge of leaving the circular path, which happens when the net force is zero.
Therefore, the largest value that the radius (r) can have so that the car remains in contact with the circular track is infinite. In other words, the loop-the-loop trick can be performed with any radius as long as the initial speed of the car is 5.0 m/s.