Circular mechanics question: finding an expression for T

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The following is an extract from my Further Mechanics 2 book (A level):

enter image description here

So I had a go at the above problem; here is what I attempted:

enter image description here

But the actual answer is apparently this:

enter image description here

Can someone please follow my workings through and point at where I might have gone wrong? (or possibly the mark scheme as this is a first edition copy)

Thanks!

asked Jul 31 at 19:02

Raihaan

203

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up vote
0
down vote

favorite

The following is an extract from my Further Mechanics 2 book (A level):

enter image description here

So I had a go at the above problem; here is what I attempted:

enter image description here

But the actual answer is apparently this:

enter image description here

Can someone please follow my workings through and point at where I might have gone wrong? (or possibly the mark scheme as this is a first edition copy)

Thanks!

asked Jul 31 at 19:02

Raihaan

203

add a commentÂ |Â

up vote
0
down vote

favorite

The following is an extract from my Further Mechanics 2 book (A level):

enter image description here

So I had a go at the above problem; here is what I attempted:

enter image description here

But the actual answer is apparently this:

enter image description here

Can someone please follow my workings through and point at where I might have gone wrong? (or possibly the mark scheme as this is a first edition copy)

Thanks!

asked Jul 31 at 19:02

Raihaan

203

The following is an extract from my Further Mechanics 2 book (A level):

enter image description here

So I had a go at the above problem; here is what I attempted:

enter image description here

But the actual answer is apparently this:

enter image description here

Can someone please follow my workings through and point at where I might have gone wrong? (or possibly the mark scheme as this is a first edition copy)

Thanks!

asked Jul 31 at 19:02

Raihaan

203

asked Jul 31 at 19:02

Raihaan

203

asked Jul 31 at 19:02

Raihaan

203

asked Jul 31 at 19:02

Raihaan

203

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

Putting the $m$ inside the square root (as in the "actual answer") is obviously wrong.

Your mistake is here:
$$
T sintheta - R costheta = frac12 T - fracsqrt32 R
neq frac12 T + fracsqrt32 R.
$$

You wrote $T sintheta - R costheta$ correctly, but later you wrote
$frac12 T + fracsqrt32 R,$ which was supposed to be the same thing but was not.

If you had written $frac12 T - fracsqrt32 R$ instead,
which is the correct set of substitutions,
I think the rest of your calculations would have come out correct.

answered Jul 31 at 21:40

David K

48k340107

Thanks, classic example of a sign error!
â€“Â Raihaan
Aug 1 at 23:32

add a commentÂ |Â

up vote
0
down vote

With $theta = arctanleft(frac 12right) = fracpi6$

Calling

$$
beginarrayrcll
delta &=& 2 r sinthetacostheta&mboxMass distance from rotation axis\
vec H &=&omega^2delta m(-1,0)&mboxCentripetal force\
vec W &=& m g(0,-1)&mboxWeigth force\
vec T &=& t(sintheta,costheta)&mboxTension in the rod\
vec R &=& r_0(-costheta,sintheta)&mboxSphere reaction force
endarray
$$

in equilibrium we have

$$
vec H+vec W+vec T+vec R = vec 0
$$

$$
left{
beginarrayrcl
-2 m r_0costheta sintheta omega ^2-r costheta+t sintheta&=&0 \
-g m+t costheta+r sintheta&=&0 \
endarray
right.
$$

solving for $t,r_0$ we have

$$
t=frac14 sqrt3 m left(2 g+omega ^2 rright)\
r_0 = frac14 m left(2 g-3 omega ^2 rright)
$$

The supposed answer with $sqrtm$ cannot be correct. It is not dimensionallly feasible.

edited Aug 1 at 13:08

answered Jul 31 at 21:26

Cesareo

5,5612412

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

accepted

Putting the $m$ inside the square root (as in the "actual answer") is obviously wrong.

Your mistake is here:
$$
T sintheta - R costheta = frac12 T - fracsqrt32 R
neq frac12 T + fracsqrt32 R.
$$

You wrote $T sintheta - R costheta$ correctly, but later you wrote
$frac12 T + fracsqrt32 R,$ which was supposed to be the same thing but was not.

If you had written $frac12 T - fracsqrt32 R$ instead,
which is the correct set of substitutions,
I think the rest of your calculations would have come out correct.

answered Jul 31 at 21:40

David K

48k340107

Thanks, classic example of a sign error!
â€“Â Raihaan
Aug 1 at 23:32

add a commentÂ |Â

up vote
1
down vote

accepted

Putting the $m$ inside the square root (as in the "actual answer") is obviously wrong.

Your mistake is here:
$$
T sintheta - R costheta = frac12 T - fracsqrt32 R
neq frac12 T + fracsqrt32 R.
$$

You wrote $T sintheta - R costheta$ correctly, but later you wrote
$frac12 T + fracsqrt32 R,$ which was supposed to be the same thing but was not.

If you had written $frac12 T - fracsqrt32 R$ instead,
which is the correct set of substitutions,
I think the rest of your calculations would have come out correct.

answered Jul 31 at 21:40

David K

48k340107

Thanks, classic example of a sign error!
â€“Â Raihaan
Aug 1 at 23:32

add a commentÂ |Â

up vote
1
down vote

accepted

Putting the $m$ inside the square root (as in the "actual answer") is obviously wrong.

Your mistake is here:
$$
T sintheta - R costheta = frac12 T - fracsqrt32 R
neq frac12 T + fracsqrt32 R.
$$

You wrote $T sintheta - R costheta$ correctly, but later you wrote
$frac12 T + fracsqrt32 R,$ which was supposed to be the same thing but was not.

If you had written $frac12 T - fracsqrt32 R$ instead,
which is the correct set of substitutions,
I think the rest of your calculations would have come out correct.

answered Jul 31 at 21:40

David K

48k340107

Putting the $m$ inside the square root (as in the "actual answer") is obviously wrong.

Your mistake is here:
$$
T sintheta - R costheta = frac12 T - fracsqrt32 R
neq frac12 T + fracsqrt32 R.
$$

You wrote $T sintheta - R costheta$ correctly, but later you wrote
$frac12 T + fracsqrt32 R,$ which was supposed to be the same thing but was not.

If you had written $frac12 T - fracsqrt32 R$ instead,
which is the correct set of substitutions,
I think the rest of your calculations would have come out correct.

answered Jul 31 at 21:40

David K

48k340107

answered Jul 31 at 21:40

David K

48k340107

answered Jul 31 at 21:40

David K

48k340107

answered Jul 31 at 21:40

David K

48k340107

Thanks, classic example of a sign error!
â€“Â Raihaan
Aug 1 at 23:32

add a commentÂ |Â

Thanks, classic example of a sign error!
â€“Â Raihaan
Aug 1 at 23:32

Thanks, classic example of a sign error!
â€“Â Raihaan
Aug 1 at 23:32

add a commentÂ |Â

up vote
0
down vote

With $theta = arctanleft(frac 12right) = fracpi6$

Calling

in equilibrium we have

$$
vec H+vec W+vec T+vec R = vec 0
$$

$$
left{
beginarrayrcl
-2 m r_0costheta sintheta omega ^2-r costheta+t sintheta&=&0 \
-g m+t costheta+r sintheta&=&0 \
endarray
right.
$$

solving for $t,r_0$ we have

$$
t=frac14 sqrt3 m left(2 g+omega ^2 rright)\
r_0 = frac14 m left(2 g-3 omega ^2 rright)
$$

The supposed answer with $sqrtm$ cannot be correct. It is not dimensionallly feasible.

edited Aug 1 at 13:08

answered Jul 31 at 21:26

Cesareo

5,5612412

add a commentÂ |Â

up vote
0
down vote

With $theta = arctanleft(frac 12right) = fracpi6$

Calling

in equilibrium we have

$$
vec H+vec W+vec T+vec R = vec 0
$$

$$
left{
beginarrayrcl
-2 m r_0costheta sintheta omega ^2-r costheta+t sintheta&=&0 \
-g m+t costheta+r sintheta&=&0 \
endarray
right.
$$

solving for $t,r_0$ we have

$$
t=frac14 sqrt3 m left(2 g+omega ^2 rright)\
r_0 = frac14 m left(2 g-3 omega ^2 rright)
$$

The supposed answer with $sqrtm$ cannot be correct. It is not dimensionallly feasible.

edited Aug 1 at 13:08

answered Jul 31 at 21:26

Cesareo

5,5612412

add a commentÂ |Â

up vote
0
down vote

With $theta = arctanleft(frac 12right) = fracpi6$

Calling

in equilibrium we have

$$
vec H+vec W+vec T+vec R = vec 0
$$

$$
left{
beginarrayrcl
-2 m r_0costheta sintheta omega ^2-r costheta+t sintheta&=&0 \
-g m+t costheta+r sintheta&=&0 \
endarray
right.
$$

solving for $t,r_0$ we have

$$
t=frac14 sqrt3 m left(2 g+omega ^2 rright)\
r_0 = frac14 m left(2 g-3 omega ^2 rright)
$$

The supposed answer with $sqrtm$ cannot be correct. It is not dimensionallly feasible.

edited Aug 1 at 13:08

answered Jul 31 at 21:26

Cesareo

5,5612412

With $theta = arctanleft(frac 12right) = fracpi6$

Calling

in equilibrium we have

$$
vec H+vec W+vec T+vec R = vec 0
$$

$$
left{
beginarrayrcl
-2 m r_0costheta sintheta omega ^2-r costheta+t sintheta&=&0 \
-g m+t costheta+r sintheta&=&0 \
endarray
right.
$$

solving for $t,r_0$ we have

$$
t=frac14 sqrt3 m left(2 g+omega ^2 rright)\
r_0 = frac14 m left(2 g-3 omega ^2 rright)
$$

The supposed answer with $sqrtm$ cannot be correct. It is not dimensionallly feasible.

edited Aug 1 at 13:08

answered Jul 31 at 21:26

Cesareo

5,5612412

edited Aug 1 at 13:08

answered Jul 31 at 21:26

Cesareo

5,5612412

answered Jul 31 at 21:26

Cesareo

5,5612412

answered Jul 31 at 21:26

Cesareo

5,5612412

add a commentÂ |Â

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