You want to show your friend what you have learned so far about circular motion, so you take a bucket of water and start swinging it in a vertical circle. The water remains in the bucket even when the bucket is upside down at the top of the circle. As you are swinging it, you become distracted as your mom calls you to come inside, and the speed of the bucket of water decreases. Splash! The bucket of water reached the top position on the circle and the water dumped out, getting you all wet. Why did the water stay in the bucket at first and then fall out when the speed decreased? When the speed decreased, it went below the critical speed required to keep the water moving in circular motion. This article will discuss how the critical speed applies to objects experiencing vertical circular motion.
Definition of Critical Speed in Circular Motion
An object in circular motion has linear velocity in a direction tangent to its circular path. For an object moving in non-uniform circular motion, the magnitude of its velocity, or its speed, changes as it moves along the path. This is what occurs when an object, like a ball or a bucket of water, is swung in a vertical circular path because the force of gravity is always acting downward on the object. The critical speed is the minimum speed that the object must have at the top position of a vertical circular path in order to complete the circular path.
The critical speed is the minimum speed that an object must have at the top position of a vertical circular path in order to complete the circular path when there is no upward support.
Let’s use the example of a ball on a string being swung in a vertical circular path. In the article, "Circular Motion", we identified the two forces acting on the ball as the force of gravity and the tension force from the string. The force of gravity is dependent on the mass of the ball, which remains the same, so it is constant while you swing the ball. The amount of tension, on the other hand, changes depending on how quickly you swing the ball. The faster you swing it, the more tension there is in the string. This makes sense because we know from Newton’s second law that the centripetal force, the sum of the tension and radial component of gravity in this case, is proportional to the linear velocity squared. If you do not swing the ball fast enough, the string will have no tension and the ball will fall out of circular motion and move in a parabolic motion as a projectile instead.
At the top of a vertical circle, a ball swung on a string will remain in circular motion if swung with at least the critical speed, StudySmarter Originals
When the ball is at the top position on the circle, as shown above, and if the speed of the ball is at least the critical speed, it will continue moving in a circular path. If it is less than the critical speed, the string will have no tension, and the ball with fall. Another example of this is a rollercoaster car going around a circular loop. The rollercoaster car will complete a circular loop if it has at least the critical speed at the top of the loop. If it were to have less than the critical speed at the top of the loop, the rollercoaster car would fall down (if it weren't attached to the rails). In the next section, we will discuss how to calculate the critical speed.
Critical Speed Calculation
Consider again the ball being swung on a string at the top position of the circular path. We have already identified the forces acting on the ball as tension and gravity. At this location, the force of gravity points down, or radially inward, so it contributes to the centripetal force along with tension:
\[\begin{align}F_\text{c}&=ma_\text{c},\\ F_T+F_g&=m\frac{v^2}{r}.\end{align}\]
In this equation, \(m\) is the mass of the ball, \(v\) is the linear velocity, and \(r\) is the radius of the circular path. Since we are looking for the critical speed, we need to calculate the minimum velocity of the ball at the top position. The minimum velocity of the ball at this location occurs when the tension force is zero. Also, remember that the force of gravity is the mass multiplied by the acceleration due to gravity, \(F_g=mg\). Making these substitutions gives us:
\[\begin{align}0+F_g&=m\frac{v^2}{r},\\mg&=m\frac{v^2}{r},\\gr&=v^2,\\v&=\sqrt{gr}.\end{align}\]
This is the minimum velocity of the ball at the top of the circle such that it doesn't leave its circular trajectory.
Formula for Critical Speed
The derivation above led us to the formula for the critical speed of an object in vertical circular motion:
\[v_\text{c}=\sqrt{gr}.\]
Notice how the critical speed depends only on the acceleration due to gravity and the radius of the circular path, not on the mass of the object.
Critical Speed Examples
As mentioned above, the equation for the critical speed can be used for any object experiencing vertical circular motion. Let’s do a couple of examples to get practice with finding the critical speed.
A bucket of water is being swung in a vertical circular path with a radius of \(1.5\,\mathrm{m}\). Find the critical speed that the bucket of water must have so that the water will not spill.
We are looking for the minimum speed of the bucket of water at the top of the circular path, so we can use the equation for critical speed. The critical speed in order to keep the water from spilling is:
\[v_\text{c}=\sqrt{gr}=\sqrt{9.81\,\frac{\mathrm{m}}{\mathrm{s}^2}\cdot1.5\,\mathrm{m}}=3.8\,\frac{\mathrm{m}}{\mathrm{s}}.\]
A \(65\,\mathrm{kg}\) passenger is riding a rollercoaster that goes around a vertical loop of radius \(15\,\mathrm{m}\). If the rollercoaster is moving at least at the critical speed, the passenger will remain in the seat when upside down at the top of the loop without the seatbelt applying any force. If it is traveling at a speed less than the critical speed, the seatbelt will need to apply force to keep the passenger in place. What is the critical speed at the top of the loop? If the rollercoaster travels at a speed 1.5 times greater than the critical speed, what is the magnitude of the normal force from the seat on the passenger?
At the top of the vertical loop, the normal force from the seat and the force of gravity act downwards, or radially inward, on the passenger, as shown in the image above. So, our equation of motion is:
\[\begin{align}F_\text{c}&=ma_\text{c},\\F_N+mg&=m\frac{v^2}{r}.\end{align}\]
To find the critical speed, we set the normal force to zero, as we did with the tension before, and we arrive at the same equation for the critical speed:
\[\begin{align}0+mg&=m\frac{v_\text{c}^2}{r},\\v_\text{c}&=\sqrt{gr}.\end{align}\]
So the critical speed of the passenger is \(v_\text{c}=\sqrt{9.81\,\frac{\mathrm{m}}{\mathrm{s}^2}\cdot15\,\mathrm{m}}=12\,\frac{\mathrm{m}}{\mathrm{s}}\).
When the rollercoaster travels at a speed 1.5 times greater than the critical speed, we have \(v=1.5v_\text{c}\). We get the normal force by substituting this into the equation we found above:
\[\begin{align}F_N+mg&=m\frac{v^2}{r},\\F_N&=m\left(\frac{\left(1.5v_\text{c}\right)^2}{r}-g\right),\\F_N&=m\left(\frac{2.25gr}{r}-g\right),\\F_N&=1.25mg,\\F_N&=8.0\cdot 10^2\,\mathrm{N}.\end{align}\]
Critical Speed - Key takeaways
- The critical speed is the minimum speed that an object must have at the top position of a vertical circular path in order to remain in circular motion.
- The centripetal force acting on an object is proportional to the square of the linear velocity.
- The formula for the critical speed of an object is \(v_\text{c}=\sqrt{gr}\).
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