#1
Hi,

Can anyone help with questions 2b and 3?

For 2b I get negative answers, and surely r dot isn't negative as velocity can't be negative.

Thanks!
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2 years ago
#2
(Original post by casiocrate)
Hi,

Can anyone help with questions 2b and 3?

For 2b I get negative answers, and surely r dot isn't negative as velocity can't be negative.

Thanks!
2b. r dot will be negative until the mass reaches position X - see diagram.

3. Most of the information given is redundant for working out the relative velocities.

What's the velocity of A? And of B? You may find it useful to slap in a coordinate system - standard Cartesian.

Then for relative velocities:
From A's point of view, velocity B - velocity A.

In a similar fashion for B's.
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2 years ago
#3
(Original post by ghostwalker)
...
You're back!
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#4
(Original post by ghostwalker)
2b. r dot will be negative until the mass reaches position X - see diagram.

3. Most of the information given is redundant for working out the relative velocities.

What's the velocity of A? And of B? You may find it useful to slap in a coordinate system - standard Cartesian.

Then for relative velocities:
From A's point of view, velocity B - velocity A.

In a similar fashion for B's.

For Q3, my professor gave a hint about reading about angular velocity. I understand that car B is moving on a circular path, but I'm not sure how I should incorporate it here.
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2 years ago
#5
(Original post by casiocrate)
For Q3, my professor gave a hint about reading about angular velocity. I understand that car B is moving on a circular path, but I'm not sure how I should incorporate it here.
I assumed that there was more to the question than you'd posted and as such were only interested in the relative velocites at the instance given, for this part.

However, if you're interested in the relative velocites at any time:

If we have the angular velocity, we can work out the position and hence velocity (differentiate) of car B at any time, t say. And then proceed as before.
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#6
Well I'm not sure what is wanted exactly, the question was given as above and the hint was passed on to me after I'd posted, so now I am slightly confused, but I'll give the second suggestion a try (I'm not really sure how though, I struggle with mechanics).
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2 years ago
#7
(Original post by casiocrate)
Well I'm not sure what is wanted exactly, the question was given as above and the hint was passed on to me after I'd posted, so now I am slightly confused, but I'll give the second suggestion a try (I'm not really sure how though, I struggle with mechanics).
I'd start by assuming at t=0 the position is as per diagram.

Then, knowing the speed of car B, we can work out it's angular velocity. Hence we can work out the angle through which it's turned by time t, and get it's position....
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#8
(Original post by ghostwalker)
I'd start by assuming at t=0 the position is as per diagram.

Then, knowing the speed of car B, we can work out it's angular velocity. Hence we can work out the angle through which it's turned by time t, and get it's position....
Thanks.

So here's where I got up to. It's probably wrong.

At t=0s, r(A) = -100i r(B) = 70j (not really sure where/how I use these)
v(A) = 20i m/s
Angular velocity = linear velocity/radius of circular path = 35/70 = 0.5 rad/s

r(B) = xi + yj where x = rcos(wt) and y = rsin(wt) so r = rcos(wt)i + rsin(wt)j

Differentiating:

v(B) = rwcos(wt)j - rwsin(wt)i = 35cos(0.5t)j - 35sin(0.5t)i

Then v(A/B) = v(A) - v(B) = 20i - (35cos(0.5t)j - 35sin(0.5t)i) = (20 + 35sin(0.5t)i -35cos(0.5t)j m/s

v(B/A) = v(B)- v(A) = 35cos(0.5t)j - 35sin(0.5t)i - 20i = 35cos(0.5t)j - (35sin(0.5t) + 20)i m/s

EDIT: would v(A/B) actually be v(A) + v(B) since v(B) is moving in the opposite direction? and similar for v(B/A)?
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2 years ago
#9
(Original post by casiocrate)
Thanks.

So here's where I got up to. It's probably wrong.

At t=0s, r(A) = -100i r(B) = 70j (not really sure where/how I use these)
v(A) = 20i m/s
Angular velocity = linear velocity/radius of circular path = 35/70 = 0.5 rad/s

r(B) = xi + yj where x = rcos(wt) and y = rsin(wt) so r = rcos(wt)i + rsin(wt)j
At the starting position OB is perpendicular to OA, and the angle is measured counterclockwise from that initial line, i.e. OB.

So, r(B)= - rsin(wt)i + rcos(wt)j

Differentiating:

v(B) = rwcos(wt)j - rwsin(wt)i = 35cos(0.5t)j - 35sin(0.5t)i

Then v(A/B) = v(A) - v(B) = 20i - (35cos(0.5t)j - 35sin(0.5t)i) = (20 + 35sin(0.5t)i -35cos(0.5t)j m/s

v(B/A) = v(B)- v(A) = 35cos(0.5t)j - 35sin(0.5t)i - 20i = 35cos(0.5t)j - (35sin(0.5t) + 20)i m/s
The idea is sound, but see my previous comment.

EDIT: would v(A/B) actually be v(A) + v(B) since v(B) is moving in the opposite direction? and similar for v(B/A)?
Since we're dealing with vector quantities, the direction is catered for, so:

v(A/B) = v(A) - v(B)

This will always be the case.
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#10
(Original post by ghostwalker)
At the starting position OB is perpendicular to OA, and the angle is measured counterclockwise from that initial line, i.e. OB.

So, r(B)= - rsin(wt)i + rcos(wt)j

The idea is sound, but see my previous comment.

Since we're dealing with vector quantities, the direction is catered for, so:

v(A/B) = v(A) - v(B)

This will always be the case.
Thanks!I ended up with the same result as using v(A) = 20i and v(B) = -35i when t = 0 (v(A/B) = 55i m/s and v(B/A) = -55i m/s), so I'm guessing I formulated the position vector for B correctly(?).
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2 years ago
#11
(Original post by casiocrate)
Thanks!I ended up with the same result as using v(A) = 20i and v(B) = -35i when t = 0 (v(A/B) = 55i m/s and v(B/A) = -55i m/s), so I'm guessing I formulated the position vector for B correctly(?).
If you're following my previous post, I'd agree.
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