Centripetal Definition and 1000 Threads

Let's say we have an object moving in a circular path with radius R and rough surface. We want to know the maximum distance covered by the car before it slip. Should we use the resultant of both the tangential acceleration and the radial acceleration or just one of them to put in the...

I have some questions... Firstly I drew the FBD of the said block at an angle theta from the vertical Which force causes the block to lose contact.ie the normal to become zero... Is it because the forces in the horizontal direction surpass some limit value of the net force in the radial...

(a) I use centripetal force to solve this question, even though the question states elliptical orbit $$\frac{GMm}{r^2}=m\omega^2 r$$ $$\omega^2 r^3=GM$$ So, $$\omega_{A}^{2} r_{A}^{3}=\omega_{B}^{2} r_{B}^{3}$$ $$\frac{r_A}{r_B}=\left(\frac{\omega_{B}}{\omega_{A}}\right)^{\frac{2}{3}}$$ Is...

I don't understand why am I getting opposite answers. I get 250N for the lowest point and 750 for the highest point. For the highest point: ##F_{net}=F_{cp}+F_{mg}-F_N## and then ##F_N=ma_{cp}+mg = 50*(25/5)+500=750N## because I've been told that ##F_{cp}## always acts towards the center of the...

I calculated the acceleration which is 0.804m/s^2. From there I calculated the centripetal force which is 0.402N. I think my lack of answer is due to my lack of understanding of the concept of what the centripetal force is at the top of the circle. Would it not be Fc = Fg - Ft as the ball...

I had an exam ques which was as follows: The first part is clear to me.....it is uniform (or constant) speed. I am in doubt on the account of the second part as the answer key says this: So the overall question concerning the second part is as follows: We know that the direction of...

So far: I an having trouble in the FBD. I drew one completely opposite to one I found on google . in this image NORmal force is pointed away from the cyclist and centripetal force is pointed away from the centre . mine was the complete opposite am i wrong?

The acceleration near the earth, due to the force of gravity is g. Now every particle when moving in a curve trajectory had a centripetal acceleration towards the center (say the sun) a=(v^2)/R. If this is true why we measure weight only with the account of g? I guess when R is big it might be...

Surface acceleration is proportional to density and radius of planet (as 2 powers of R cancel with the volume) g(moon)/g(earth) = density(moon)*radius(moon)/density (earth)*radius(earth) = (1/4)*density(moon)/density(earth)

Supposed answer = 0.482n

See attached image. The solution to this problem calculates v2 at the top of the roller coaster ride. Why is that? Shouldn't you calculate v2 at the bottom of the roller coaster ride as you require the maximum velocity there to get around the loop?

Unsure

I have attempted to solve for the velocity by setting the centripetal force (mv2)/r to the normal force pointed to the center of rotation (mg). This approach seems to give the incorrect solution and I am unsure of my misunderstandings.

Relevant formulae:- Angular velocity in uniform circular motion ##=## ##\omega## ##=## ##\frac {2\pi} t##, where ##t## is the time taken to complete one revolution. Centripetal acceleration in uniform circular motion ##=## ##a## ##=## ##\omega^2r##, where ##r## is the radius of the circular...

Hello, I am attempting to correctly solve this problem, however I end up with an equation that is slightly different as the one provided in the textbook solution. For question (a) I get the same thing, just instead of cos, I have cos^2 and I can't figure out where I went wrong. My process was...

Hi, I just had a question about this homework question. I am not given the mass at all in any portion of the question. Fs = Fc because the static friction is the thing that keeps the rider stuck to the wall My answer came out to about 3.4 m/s for the minimum speed that keeps the rider stuck to...

To be honest I am a bit clueless first with how to interrupt this question I think the bead is going around a wall type thing where there is friction both in the up and X direction. Some hint to get some ideas running would be great

How come the solution requires you to do mv^2/4 when the formula is just v^2 /r?

As I understand it, when a body undergoes uniform circular motion its velocity does not change in magnitude but instead direction. This change in velocity, or acceleration, is directed inward towards the center of the circle. If a body was not experiencing a net centripetal acceleration, then...

Hello, as you can see i am trying to understand conceptually how the tires during turning create a centripetal force. It was explained to me that as we turn the car tires, the tires similar to a ski or a wedge, now want to push the ground to the side and forward. If the ground was loose, this...

I am confused. See my diagram below. With the Earth rotating, I think the scale would read the force of gravity or 448.4 N. If the Earth were not rotating, would the scale read less or more due to the effect of centripetal force? I tend to think more by an amount of 1.78 N. Is this correct? If...

So for this problem the solution used Cartesian coordinates but I was wondering wouldn’t it be easier to use Normal and tangential coordinate because the bar is undergoing centripetal motion? Also on the right diagram shouldn’t the acceleration be down and not up. The reason I think that is...

Let me imagine a box placed on a table. It has got no acceleration. If I were a person who trusted Newton's laws then I would argue that the net force on the box should be zero. Now in another situation I am an observer outside the Earth and I see that the box is rotating along with the earth...

Ok, so we know that if one were inside a donut-shaped spaceship that is rotating around it's axis, that the passengers will experience centripetal force. It seems obvious to say that the ship is rotating relative to the nearby stars and planets. So far, so good. But... what if we removed all...

For the displacement, how do I figure out the angle theta between the points? And how does the speed at which the string retracts affect the centripetal force?

(a) Using COE, $$mgh = 0.5mv^2 + 0.5I\omega^2$$ I solved it, where $$\omega = 112 rad/s$$ (b) This is the part where I have question or problem. I saw my course mate working and he start of with finding centripetal acceleration. $$a_c = \frac{v^2}{r} = \frac{(r_0\omega)^2}{R_0}$$ Why isn't it...

Hi, Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of...

Hi, A mother pushes her child on a swing so that his speed is 9.00 m/s at the lowest point of his path. The swing is suspended 2.00 m above the child’s center of mass. (a) What is the magnitude of the centripetal acceleration of the child at the low point? (b) What is the magnitude of...

Yeah, as I said, i have no idea why we can assume beta to be the same for any position at given alpha.

Summary:: Just want to know if I'm on the right track with this question. Hi, so this is what I have for my assignment: A washing machines drum is rotating rapidly about a vertical axis (a so-called toploader). A wet sock is stuck on the inside, halfway up the drum, and the drum begins to slow...

Ff = Fc (mv^2)/r = kmg (v^2)/r = kg v^2 = kgr v^2 = 0.7*9.8*200 v = 37.04 m/s I chose option e, and its wrong. What am i doing wrong here?

1. Newton's Second Law states F=ma and the formula for centripetal acceleration is v^2/r Therefore, F= mv^2/r Would this be complete, I just feel that I should need to do something further but I am not sure what? 2.F=mv^2/r Gravitational force = GMm/r^2 Gravity is the cause of centripetal...

Given such a diagram as shown above, we know that the normal force must be mg/sintheta. How is this normal force greater than the gravitational force conceptually? Is it due to the horizontal traveling (which must have been started by someone exerting a force?) compressing the sides of the cone...

Situation: Let’s say we have a wire bent into a circular shape, there lies a bead through the wire and it can slide through it. The wire is kept in vertical plane and is swung along the axis AB. My question : How the centripetal force is provided to the bead? The bead will go into a...

Hey guys, Theres something I've been confused about when looking at circular motion. When does an object have just centripetal acceleration as the acceleration of the object, if ever. I think that the acceleration vector is between the centripetal and tangential acceleration when an objects...

Centripetal force is defined as the force causing the body to follow a curved path, acting towards the center and always orthogonal to the direction of motion. For uniform circular motion the formula for centripetal acceleration is $$a_c = \frac{v^2}{r}$$. But my understanding of centripetal...

1. When a car turns there is a centripetal force towards the centre. This centripetal force is labelled as a static frictional force. I don't understand where this static frictional force arises from. Friction is meant to oppose motion, but I don't see the motion that is parallel to the friction...

My initial attempt: Total Centripetal force on the cylinder would be given by $$\textbf{F}_{net} = mR\omega^2 \textbf{e}_1+mr_{cm}\omega^2 \textbf{e}_2$$ where the vectors e_1 and e_2 have magnitude 1 and point radially outwards (and continuously changing as the cylinder rolls down) as marked in...

I believe I solved this. Is this solution true? Can please anyone just check?

I think I have solved the first three, and only really need help on question four. For number one, I used Fc = (Mv^2)/R and just rearranged it for velocity so I ended up with v = sqrt(ac * R) For number 2 I used Ff = Fn*mu and got Mg*mu = Ff For number 3 I used w = Ff*d and got w = -Mg*mu*l...

Well i think centripetal force is inversely proportional to the radius, so I'm guessing it would decrease but not sure to be honest.

So far what we know about the circular motion is that an object moving in a circle experiences a force towards the center of the circle and as a result accelerates towards this center. But we also know that an object always moves in the direction of resultant force - if two tractors moving at...

I tried this problem 3 times. I only have two attempts left. First time: Centripetal acceleration: 7560 m/s^2 Centripetal Force: 4.7 Newtons Second time: Centripetal acceleration: 25.032 Centripetal Force: 4.7 Third Time: Centripetal...

A demonstration of the direction water swirls in two hemispheres Sync the videos yourself: http://toiletswirl.com For the record Destin and I repeat... 2 different Videos synced. The explanation of why water swirls in a specific direction is at about 4 minutes into the video.

Matt and Hugh play with a tennis ball and a brick. Then they do some working out to derive the formula for the centripetal force (a = v^2/r) by differentiati...

Suppose, there is an object in a circular path that goes with a cirtain speed. What happens, if suddenly the centripetal force increases? a) The object remains in the path but its speed increases b) The object exits the circular path c) Any other situation Please, explain your answer, thanks

According to the Newton's third law "For every action, there is an equal and opposite reaction." When a car (or a bike) turns, How does the car (bike) exert force outward (in the opposite direction of centripetal friction force)?

So here is what is going on in my mind: We have a turn that is 400m away from the center of the turn. The faster the car goes, the harder it is for it to maintain its radius. We have a component of the normal force that points towards the center, and static friction which does the same. I...

Correct me if I'm wrong, but I think it is not possible to solve (1) with all the data that's given. As for (2), I have come up with the following solutions: (a) - The tension in the string acts as the centripetal force on the fuzzy dice (b) - The frictional force between the road and the car...

I am not too sure as to how to approach part c. of this question. In the vertical plane, the centripetal force is provided by the normal force and the force of gravity. However, the solution to this problem includes a description of the forces at the top of the loop, where the normal force is...