How do you know when to use angular velocity (w) as opposed to finding a velocity using the circumference ( that would be in m/s) ?

One can use either. You can convert angular velocity to velocity if you know the radius, There isn't a rule as such. One chooses based on the data given and the ease of solving the problem. For example, if the data specifies velocity in radian/sec or revolutions/sec, using the angular velocity formula is probably the better choice as you avoid the hassle of all the mathematics involved in conversion of velocities. Similarly, if the data specifies velocity as meters/sec, go for the usual velocity formula.

I do understand the solution of exercise 1, but something else about it confuses me. the force of gravitation should be acting on m1 as well, right? so why is m1 not accelerating downwards? which force balances the force of gravitation out?

I think you are supposed to imagine that the ball is on a table with a hole drilled in the table. Otherwise the rope going from m1 to the tube would be at an angle and there would be two components of force applying tension to the rope, and m2 would have to be large enough not only to provide the centripetal force but also to provide the upward force to offset m1's weight.

why does the m1rw^2 = m2g? i don't understand how both equal?

With physics problems, they do this 'massless string approximation' where you ignore the mass of the string (since in most cases it is much smaller than the masses attached) and so you can treat the two objects connected to the string as directly interacting with each other. It follows that the tension in the string must therefore be the same at all points, and so the tension (you can find from mass 2, which is m2*g) is equal to the centripetal force causing mass 1 to spin around. If there is anything you need me to clarify feel free to ask!

why does the centripetal force equal to the tension force in some cases?

Centripetal force, for an object in circular motion, is just the force maintaining the circular motion, the force that prevents the object from flying away tangentially. Unless this force is present, no object can execute a pure circular motion, because in absence of a force, a body moves uninterrupted along a straight line (Newton's first law). In most scenarios you see in physics' problems, a string is the provider of the centripetal force; it acts as the tether to the object making it go in a circle rather than flying off straight. In such a scenario, the force acting on the object (centripetal force) is the tension in the string. They are the same things.

In Exercise 1, if the ball was made to rotate faster, would the system (of the ball and mass m2) start accelerating upwards?

Sure, why not? To see this, imagine what happens if the ball rotates slower, or comes to a stop. m2 will fall, right? So it must rise if the opposite happens.

In the last question, why is 35 divided by 2? Thanks!

Since each blade is length 35 meters, to find the radius we divide by two. The center of mass for the radius of rotation is located in the middle of the blade, thus you divide 35 by 2 in order to find the distance from the center to the edge of the blade.

Main content

Course: Physics library > Unit 4

Lesson 2: Centripetal forces

What is a centripetal force?

Google Classroom

Learn what centripetal forces are and how to calculate them.

What is a centripetal force?

A centripetal force is a net force that acts on an object to keep it moving along a circular path.

In our article on centripetal acceleration, we learned that any object traveling along a circular path of radius

r

with velocity

v

experiences an acceleration directed toward the center of its path,

a = \frac{v^{2}}{r}

However, we should discuss how the object came to be moving along the circular path in the first place. Newton’s 1ˢᵗ law tells us that an object will continue moving along a straight path unless acted on by an external force. The external force here is the centripetal force.

It is important to understand that the centripetal force is not a fundamental force, but just a label given to the net force which causes an object to move in a circular path. The tension force in the string of a swinging tethered ball and the gravitational force keeping a satellite in orbit are both examples of centripetal forces. Multiple individual forces can even be involved as long as they add up (by vector addition) to give a net force towards the center of the circular path.

Starting with Newton's 2ⁿᵈ law :

a = \frac{F}{m}

and then equating this to the centripetal acceleration,

\frac{v^{2}}{r} = \frac{F}{m}

We can show that the centripetal force

F_{c}

has magnitude

F_{c} = \frac{m v^{2}}{r}

and is always directed towards the center of the circular path. Equivalently, if

ω

is the angular velocity then because

v = r ω

F_{c} = m r ω^{2}

Tethered ball

One apparatus that clearly illustrates the centripetal force consists of a tethered mass (

m_{1}

) swung in a horizontal circle by a lightweight string which passes through a vertical tube to a counterweight (

m_{2}

) as shown in Figure 1.

Exercise 1: If

m_{1}

is a

1 kg

mass spinning in a circle of radius

1 m

and

m_{2} = 4 kg

what is the angular velocity assuming neither mass is moving vertically and there is minimal friction between the string and tube?

When the string passes through the tube it redirects the force due to gravity acting on

m_{2}

to the horizontal plane. This is the centripetal force which allows

m_{1}

to rotate in a circle. Neither mass is moving vertically so the centripetal force must be exactly balanced.

\begin{aligned} F_{c} & = F_{g} \\ m_{1} r ω^{2} & = m_{2} g \end{aligned}

Which can be rearranged and solved for the rotational velocity,

\begin{aligned} ω & = \sqrt{\frac{m_{2} g}{m_{1} r}} \\ = \sqrt{\frac{4 kg \cdot 9.81 m / s^{2}}{1 kg \cdot 1 m}} \\ = 6.26 rad / s \end{aligned}

or about 1 revolution per second.

Car turns a corner

Exercise 2: A car turns a corner on a level street at a speed of

10 m/s

while sweeping out a circular path of radius

15 m

. What is the minimum coefficient of static friction between the tires and the ground for this car to make the turn without slipping?

The only force acting in the centripetal direction is the force of static friction

F_{s}

. So it must be true that,

\frac{v^{2}}{R} = \frac{F_{s}}{m}

Since we want the minimum possible coefficient of friction, we'll use the expression for the maximum static frictional force

F_{s} = μ_{s} F_{n}

. Using a normal force of

m g

we get,

\frac{v^{2}}{R} = \frac{μ_{s} m g}{m} = μ_{s} g

Which means that,

μ_{s} = \frac{v^{2}}{g R} = \frac{(10 m / s)^{2}}{(9.81 m / s^{2}) (15 m)}

μ_{s} = 0.680

Note: It's admittedly a little counterintuitive, but to find the minimum coefficient of static friction, you have to assume the static frictional force is at a maximum.

If the static frictional force weren't at a maximum, we could reduce the coefficient of static friction and still allow the car to make the turn. But the problem stated that we wanted the minimum possible coefficient of static friction that would prevent sliding.

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Mrinali.Kumar
Posted 8 years ago. Direct link to Mrinali.Kumar's post “why does the m1rw^2 = m2g...”
why does the m1rw^2 = m2g? i don't understand how both equal?
Button navigates to signup pageComment on Mrinali.Kumar's post “why does the m1rw^2 = m2g...”
(3 votes)
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- WharfRat
  Posted 8 years ago. Direct link to WharfRat's post “With physics problems, th...”
  With physics problems, they do this 'massless string approximation' where you ignore the mass of the string (since in most cases it is much smaller than the masses attached) and so you can treat the two objects connected to the string as directly interacting with each other. It follows that the tension in the string must therefore be the same at all points, and so the tension (you can find from mass 2, which is m2*g) is equal to the centripetal force causing mass 1 to spin around. If there is anything you need me to clarify feel free to ask!
  Comment on WharfRat's post “With physics problems, th...”
  (22 votes)
steffenbakken002
Posted 4 years ago. Direct link to steffenbakken002's post “In the first exercise wha...”
In the first exercise what force is preventing the second mass from accelerating downwards? And therefore changing the radius of the circle over time.

Are we supposed to be getting an instantaneous measurement for the angular velocity of the ball and not factor in how time would affect it?
Button navigates to signup pageComment on steffenbakken002's post “In the first exercise wha...”
(10 votes)
Answer
maggiecduncan
Posted 7 years ago. Direct link to maggiecduncan's post “How do you know when to u...”
How do you know when to use angular velocity (w) as opposed to finding a velocity using the circumference ( that would be in m/s) ?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Surya Bhushan Tripathi
  Posted 7 years ago. Direct link to Surya Bhushan Tripathi's post “One can use either. You ...”
  One can use either.
  You can convert angular velocity to velocity if you know the radius, There isn't a rule as such. One chooses based on the data given and the ease of solving the problem.
  For example, if the data specifies velocity in radian/sec or revolutions/sec, using the angular velocity formula is probably the better choice as you avoid the hassle of all the mathematics involved in conversion of velocities. Similarly, if the data specifies velocity as meters/sec, go for the usual velocity formula.
  Button navigates to signup page
  (4 votes)
Ev
Posted 6 years ago. Direct link to Ev's post “I do understand the solut...”
I do understand the solution of exercise 1, but something else about it confuses me. the force of gravitation should be acting on m1 as well, right? so why is m1 not accelerating downwards? which force balances the force of gravitation out?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Andrew M
  Posted 6 years ago. Direct link to Andrew M's post “I think you are supposed ...”
  I think you are supposed to imagine that the ball is on a table with a hole drilled in the table. Otherwise the rope going from m1 to the tube would be at an angle and there would be two components of force applying tension to the rope, and m2 would have to be large enough not only to provide the centripetal force but also to provide the upward force to offset m1's weight.
  Comment on Andrew M's post “I think you are supposed ...”
  (5 votes)
Richard C. Marberger
Posted 7 years ago. Direct link to Richard C. Marberger's post “Is the turbine in exercis...”
Is the turbine in exercise 2a in space? Because when the blade is pointing down there will be 98000 N from the force of gravity and when the blade is pointing up, you have to subtract 98000 N. Maybe I did not understand the problem.
Button navigates to signup pageComment on Richard C. Marberger's post “Is the turbine in exercis...”
(5 votes)
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- Jarle Mathiesen
  Posted 7 years ago. Direct link to Jarle Mathiesen's post “Yes, it seems to me like ...”
  Yes, it seems to me like the force of gravity would impact the force of tension, with the maximum tension in the downward position, and the minimal tension at the angles 0 and 180 degrees.
  Button navigates to signup page
  (0 votes)
Arav Luthra
Posted 4 years ago. Direct link to Arav Luthra's post “why does the centripetal ...”
why does the centripetal force equal to the tension force in some cases?
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(3 votes)
Answer
- Satwik Pasani
  Posted 4 years ago. Direct link to Satwik Pasani's post “Centripetal force, for an...”
  Centripetal force, for an object in circular motion, is just the force maintaining the circular motion, the force that prevents the object from flying away tangentially. Unless this force is present, no object can execute a pure circular motion, because in absence of a force, a body moves uninterrupted along a straight line (Newton's first law).
  
  In most scenarios you see in physics' problems, a string is the provider of the centripetal force; it acts as the tether to the object making it go in a circle rather than flying off straight. In such a scenario, the force acting on the object (centripetal force) is the tension in the string. They are the same things.
  Button navigates to signup page
  (2 votes)
Shreyas
Posted 7 years ago. Direct link to Shreyas's post “In Exercise 1, if the bal...”
In Exercise 1, if the ball was made to rotate faster, would the system (of the ball and mass m2) start accelerating upwards?
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(2 votes)
Answer
- Andrew M
  Posted 7 years ago. Direct link to Andrew M's post “Sure, why not? To see thi...”
  Sure, why not?
  To see this, imagine what happens if the ball rotates slower, or comes to a stop. m2 will fall, right? So it must rise if the opposite happens.
  Comment on Andrew M's post “Sure, why not? To see thi...”
  (3 votes)
Lilah Bleich
Posted 8 years ago. Direct link to Lilah Bleich's post “In the last example exerc...”
In the last example exercise in the solution where did the unit rad disappeared?
(1⋅104 kg)⋅(2.1 rad/s)2⋅(352 m)≃7.7⋅105 N
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(3 votes)
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caught.in.flight.98
Posted 8 years ago. Direct link to caught.in.flight.98's post “In the last question, why...”
In the last question, why is 35 divided by 2? Thanks!
Button navigates to signup pageComment on caught.in.flight.98's post “In the last question, why...”
(1 vote)
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- jjain654
  Posted 8 years ago. Direct link to jjain654's post “Since each blade is lengt...”
  Since each blade is length 35 meters, to find the radius we divide by two. The center of mass for the radius of rotation is located in the middle of the blade, thus you divide 35 by 2 in order to find the distance from the center to the edge of the blade.
  Button navigates to signup page
  (5 votes)
Andyunderwood12
Posted 7 years ago. Direct link to Andyunderwood12's post “The second exercise state...”
The second exercise stated that the blades weighed 10,000 Kg. Weight is equal to mass times acceleration (gravity). Kg is a weight symbol as well as a mass symbol. Why would you not take gravity in consideration for the blades weighing 10000 Kg??
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(2 votes)
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