1,000,000 Question Thread :D

[Aden]

I understand it now, it looks like the diagram attached.

Once you've found that the resultant force is 20N through pythagoras, the question becomes simple:

(a)



Angle:



Try doing (b) now

[Studyinghard]

I understand it now, it looks like the diagram attached.

Once you've found that the resultant force is 20N through pythagoras, the question becomes simple:

(a)



Angle:



Try doing (b) now

Oh thats it. I was confused with the diagram

Okay so with that in mind its simple.



yes?

[Aden]

Yes.

[Studyinghard]

<3. thanks

[kenhung123]

What is the reason that there is 0 linear acceleration a the pit and peak of a roller coaster?

[Studyinghard]

What is the reason that there is 0 linear acceleration a the pit and peak of a roller coaster?

Just a random guess here from some google-ing.

I suppose linear acceleration means that the path the object is acceleration on is straight and not like circles etc. So when your at the top or bottom of  roller coaster it stops for that instant and then goes down/negative acceleration in like a U shaped instead of straight . So it isn't linear? Idk ..just throwing it out there. 

[/0]

0 linear acceleration? I don't think that occurs at the pit and peaks of rollercoasters...
At a peak, , so you accelerate down.
At a pit, , so you accelerate up.
If there was no linear acceleration at the pit/peak then you would continue at constant velocity, so it wouldn't really be a pit/peak.

[kenhung123]

Hmm this is from heinemann "At the bottom of the pipe, however, the skateboarder will be neither slowing down nor speeding up, so the acceleration is purely centripetal at this point. The same applies at the very top of a circular path.

[appianway]

Perhaps he's refering to linear acceleration being 0... in reference to total acceleration being the vector sum of linear acceleration and centripetal acceleration?

In that case, no force acts perpendicular to the centripetal force (neglecting friction and air resistance), thus no linear acceleration occurs.

[kenhung123]

What normally contributes to 'linear' acceleration?

[Akirus]

If I'm correct in understanding what you're talking about, it would be and .

[appianway]

No, I don't think that's what he means. I think he's talking about the component of acceleration directed perpendicular (at a tangent to the circle) to the centripetal force.

It'd have to be some kind of other force. For instance, a frictional force (which opposes the momentary direction of motion) would be linear at the top or bottom, because it'd be perpendicular to the other forces acting.

[Akirus]

From the textbook:

Circular motion in a vertical plane, however, is not usually uniform; this is because the speed of the body varies. An example of this is a skateboarder practising in a half-pipe. The speed of the skateboarder will increase on the way down as gravitational potential energy is converted into kinetic energy. This means there will be a linear acceleration, a1, as well as a centripetal acceleration, ac. The resultant acceleration cannot be directed towards the centre of the circular path (Figure 2.43b). At the bottom of the ‘pipe’, however, the skateboarder will be neither slowing down nor speeding up, so the acceleration is purely centripetal at this point (Figure 2.43c). The same applies at the very top of a circular path. For this reason, motion at these points is easier to analyse.

I think the book is referring to the acceleration by gravity, which in this case would be defined by:

, where is equal to the angle from the horizontal.

Thus, at the bottom of the slope,



which means the net force accounts only for the centripetal acceleration at the pit/peak. Do I have the right idea?

[kenhung123]

I think you got it

[Studyinghard]

A student leans over a cliff 32.0 m high and throws a 20 g stone vertically downwards
with a speed of 12 m s1. What is the kinetic energy of the stone when it is 10.0 m
from the base of the cliff?