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time or distance?
Posted this question once before but answer was inconclusive. It still provokes fierce arguments amongst my friends.The question is........Does the outer end of a second hand on a clock move faster than the inner end seeing as it has a greater distance to travel in the same amount of time? one guy says it does but thats impossible, someone else said it does n`t but it should , another argues that the distance factor does not come into the equation. can some one settle this argument........PLEASE
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For more on marking an answer as the "Best Answer", please visit our FAQ.Look at the distance the end moves in a minute, now look at the distance a point near the middle moves in 60 seconds. The point at the end has to travel a much greater distance in the same amount of time.
as Speed=Distance/time
the end moves faster.
If you take it to the extreme and look at the centre of the clock it just spins on the spot and doesnt actually move anywhere so its speed is zero.
as Speed=Distance/time
the end moves faster.
If you take it to the extreme and look at the centre of the clock it just spins on the spot and doesnt actually move anywhere so its speed is zero.
The rate for completing one revolution (once a minute) is the same throughout its length but the distance traveled is proportional to the distance from the axis of rotation.
On a larger scale the Earth rotates around its axis once a day (a bit more actually since its also revolving around the Sun once a year). While the North and South Poles simply rotate in place, motion at the Equator exceeds a thousand mph.
If you really want to get technical about it the second hand, the clock, the Earth, Sun and our entire galaxy are moving through space at over 500 kilometres per second. Then again looking at it from the second hands perspective it's not moving at all, it's the clock that's turning.
I don't see why this is causing such a problem for you. It's day 1 junior physics for christ sake. Same as a wheel the outer edge of the spokes have to travel a much longer distance in the same time as the bit near the hub. Same with all the hands of a clock, simple, very simple! Why do you think it's "Impossible"?
It all depends on what you mean by "how fast". Speed is measured as "velocity". The rate at which the clock hand rotates is the "angular velocity" - which is a constant for the entire length of the clock hand. However, the linear velocity at any point on the clock hand will be proportional to the distance from the centre of the clock. Obviously at the very centrew of the clock the linear velocity will be zero... by contrast the linear velocity at the most distant point on an infinitely long clock hand will be infinite!
speed = distance travelled / time taken
this is different for each part of the second hand as each part travels a different distance during one rev.
angular velocity = no. of radians / time taken
this is the same for each part of the second hand as each part travels through the same no. of radians per second
Can't really simplify further, hope this helps.
this is different for each part of the second hand as each part travels a different distance during one rev.
angular velocity = no. of radians / time taken
this is the same for each part of the second hand as each part travels through the same no. of radians per second
Can't really simplify further, hope this helps.
speed =distance/time is not exactly true .
Velocity= d/t and acceleration =velocity/time
therefore:-
the second hand travels one revolution every minute at both ends but the outside end travels a greater distance (depending on the length of the hand) so therefore the ends travel at a different velocity the outside traveling more meters per second than the inside !
Velocity= d/t and acceleration =velocity/time
therefore:-
the second hand travels one revolution every minute at both ends but the outside end travels a greater distance (depending on the length of the hand) so therefore the ends travel at a different velocity the outside traveling more meters per second than the inside !
velocity is the rate of change of distance with time .
IE V=Meters per Second or Ms to the power of minus one
Acceleration is the rate of change of velocity with time.
IE A=Meters per second per second or Ms to the power of minus two.
It's not rocket science you know ! ( Well , it is actually )
Displacement is actually fluid dynamics and is totally another kettle of fish .
IE V=Meters per Second or Ms to the power of minus one
Acceleration is the rate of change of velocity with time.
IE A=Meters per second per second or Ms to the power of minus two.
It's not rocket science you know ! ( Well , it is actually )
Displacement is actually fluid dynamics and is totally another kettle of fish .
but displacement is a vector quantity, and distance is a scalar. The same way that speed is scalar (hence rate of change of distance) and velocity a vector (rate of change of displacement). So, the velocity of an object travelling in a circle varies constantly (since velocity is speed in a given direction) even though its speed may remain the same.
I think that the problem is, we're talking about two different displacements. You're right about the fluid one, though. And it's good that you remeber something from your Physics - I wish I could say the same for some of my students!!
I think that the problem is, we're talking about two different displacements. You're right about the fluid one, though. And it's good that you remeber something from your Physics - I wish I could say the same for some of my students!!