Understanding Angular Velocity in Circular Motion

[mutineer123]

This topic is to do with circular motion..angular velocity. I have just begun this chapter in my a2 physics, and I don't really understand why the tip moves faster than the one near the centre. I understand it through calculations(using the different length of radius and then ω/angular velocity)that the tip does indeed move faster. But why? what causes it to?

 

[Jimmy]

All points in the hand make one revolution in the same amount of time. In that time, the tip traces out a circle with a larger circumference than any other part which is closer to the center. If the tip travels a longer distance in the same amount of time, then it must have been moving faster.

 

[Naty1]

good answer posted above!

another way to think about the situation is that w is the constant [if its a good clock] not v. So rotational motion [velocity and acceleration] takes a little getting used to. You'll find torque and rotational inertia both have some interesting properties as well.

 

[justwild]

A body having same linear velocity covers a specific DISTANCE in specific time and a body with angular velocity covers specific angle (not distance) per unit time.
In the example of clock the tip of hand moves faster because it has faster linear velocity but same angular velocity.
Let the distance from horizontal to the clock tip be y and that of horizontal distance(the distance between the points where y-cuts with horizontal and hand's origin be x).
Now tanθ=y/x
which implies,
θ=tan$^{-1}$y/x
so if you increase x, y should accordingly be increased in order to maintain angle θ.
So higher x higher y and lower x lower y

 

[Jimmy]

A body having same linear velocity covers a specific DISTANCE in specific time and a body with angular velocity covers specific angle (not distance) per unit time.

The tip of the clock hand- as well as other points on the hand except for the center, does cover a specific distance. However, after one revolution, the displacement is 0.

 

[justwild]

Actually I meant that they do cover a specific distance but this distance covered (at any specific time) is not equal for every points on the hand.

 

[mutineer123]

kk thank you all for your answers. I had another question. I was reading up the chapter. It said when a car moves in a circular way around a bend(With constant ω), the centripetal force is provided by friction. I cannot grasp fully this notion. How does the friction direct inwards toward the centre, shouldn't it just act opposite to the engines force at all times?

 

[cepheid]

kk thank you all for your answers. I had another question. I was reading up the chapter. It said when a car moves in a circular way around a bend(With constant ω), the centripetal force is provided by friction. I cannot grasp fully this notion. How does the friction direct inwards toward the centre, shouldn't it just act opposite to the engines force at all times?

Because of its inertia, the car just wants to move in a straight line. But if you turn the wheels, then in order for this straight line motion to continue, the tires have to slide (relative to the road). Friction prevents this sliding from happening, and redirects the motion of the car to be in the new direction in which the tires are oriented. If you drive in a circle, this friction (which prevents the tires from slipping) will always point towards the centre of that circle.

Imagine it's winter time and you're driving, and the roads are icy. You want to drive the car around a bend that turns to the left, so you turn the steering wheel to the left. But the car keeps going in the forwards direction because the tires, although they are now pointed to the left, they just slide "sideways" across the road surface, allowing the car to continue moving forward. You've lost control! It's scary. I picked a left-going bend so that at least that way you slide *off* the road and into the ditch, instead of into oncoming traffic! (Unless you are in England, I suppose).

Sorry, I got carried away...

 

[mutineer123]

Because of its inertia, the car just wants to move in a straight line. But if you turn the wheels, then in order for this straight line motion to continue, the tires have to slide (relative to the road). Friction prevents this sliding from happening, and redirects the motion of the car to be in the new direction in which the tires are oriented. If you drive in a circle, this friction (which prevents the tires from slipping) will always point towards the centre of that circle.

Imagine it's winter time and you're driving, and the roads are icy. You want to drive the car around a bend that turns to the left, so you turn the steering wheel to the left. But the car keeps going in the forwards direction because the tires, although they are now pointed to the left, they just slide "sideways" across the road surface, allowing the car to continue moving forward. You've lost control! It's scary. I picked a left-going bend so that at least that way you slide *off* the road and into the ditch, instead of into oncoming traffic! (Unless you are in England, I suppose).

Sorry, I got carried away...

If you drive in a circle, this friction (which prevents the tires from slipping) will always point towards the centre of that circle.

I still don't understand how the friction points toward the centre. look at this rough sketch i made
http://tinypic.com/view.php?pic=24np9o5&s=6
the red arrow is the force of the car and the blue arrow is the same force, after turning. Now in both these cases the friction is the brown(backward) arrow, which is the same again for both cases, so how are these frictional force acting towards the centre, they are acting at a tangent, like the engine force.

 

[mutineer123]

okay Imagine a car turning in a levelled road. The vertical forces are balance, s there is no vertical acceleration. In the horizontal component, there is a friction force(which the book says is the unbalanced force). NOw why is the engine's force ignored? If it is accounted for, and if it is equal to the frictional force, then there would be no angular acceleration right? and hence no centripetal force...

 

[azizlwl]

If you drive in a circle, this friction (which prevents the tires from slipping) will always point towards the centre of that circle.

I still don't understand how the friction points toward the centre. look at this rough sketch i made
http://tinypic.com/view.php?pic=24np9o5&s=6
the red arrow is the force of the car and the blue arrow is the same force, after turning. Now in both these cases the friction is the brown(backward) arrow, which is the same again for both cases, so how are these frictional force acting towards the centre, they are acting at a tangent, like the engine force.

2 forces required here.
One to move forward, tangent to the curve overcoming the friction between tyres and road surface.
The other is to change the direction of the car. The direction is towards the center.

 

[cepheid]

If you drive in a circle, this friction (which prevents the tires from slipping) will always point towards the centre of that circle.

I still don't understand how the friction points toward the centre. look at this rough sketch i made
http://tinypic.com/view.php?pic=24np9o5&s=6
the red arrow is the force of the car and the blue arrow is the same force, after turning. Now in both these cases the friction is the brown(backward) arrow, which is the same again for both cases, so how are these frictional force acting towards the centre, they are acting at a tangent, like the engine force.

The role of friction in the motion of a car is more complicated than for a box that you push along a horizontal surface. For the box, you're trying to make the box *slide* forwards along the surface. As a result, friction with the surface opposes this sliding, and hence is directed rearwards.

In contrast, a car doesn't slide across the road, it rolls. For a car, friction with the road is not rearwards and it does *not* impede the motion of the car. On the contrary, friction with the road is what *allows* this motion to happen. The tire grips the road, pushing backwards on it as it rotates. The road, in turn, pushes forwards on the car, propelling it forwards. I don't have time to explain this further, so Google "Rolling Without Slipping" for more details. So there is a *forwards* frictional force (RED in the diagram below) that is parallel to the tires. The diagram below depicts a (transparent) car viewed from the top down, with its front tires rotated to the left. The car is moving forwards, but we've just turned the tires to the left. So, the tires want to slide *laterally* across the road surface, but there is a frictional force (BLUE) that prevents this sliding from happen. This is the friction that I was referring to in my previous post (the friction that fails to be present when you *skid* across an icy road)

The resultant (i.e. vector sum) of these two friction forces is the GREEN vector in the diagram. As you can see, its direction is appropriate for changing the direction of the car's motion to follow the turn.