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Page 1
Edurev123
2. Motion in a Pane
2.1 A shot fired with a velocity ?? at an angle (elevation) ?? strikes a point ?? in a
horizontal plane through the point of projection. If the point ?? is receding from the
gun with velocity ?? show that the elevation must be changed to ?? where
?????? ?? ?? =?????? ?? ?? +
?? ?? ?? ?????? ??
(2009: 12 Marks)
Solution:
Let ?? be the original position of the point.
When ?? is at rest,
???? =
?? 2
sin 2?? ??
Let the elevation be changed to ?? when the point recedes to hit it and let the point be hit
at ?? .
Then
???? =
?? 2
sin 2?? ?? ?? h?? ???????? ???????????? ???? ???????????? =
?? 2
sin 2?? ?? ·?? cos ?? =
2?? sin ?? ??
? So the point must have travelled ???? during this time.
Page 2
Edurev123
2. Motion in a Pane
2.1 A shot fired with a velocity ?? at an angle (elevation) ?? strikes a point ?? in a
horizontal plane through the point of projection. If the point ?? is receding from the
gun with velocity ?? show that the elevation must be changed to ?? where
?????? ?? ?? =?????? ?? ?? +
?? ?? ?? ?????? ??
(2009: 12 Marks)
Solution:
Let ?? be the original position of the point.
When ?? is at rest,
???? =
?? 2
sin 2?? ??
Let the elevation be changed to ?? when the point recedes to hit it and let the point be hit
at ?? .
Then
???? =
?? 2
sin 2?? ?? ?? h?? ???????? ???????????? ???? ???????????? =
?? 2
sin 2?? ?? ·?? cos ?? =
2?? sin ?? ??
? So the point must have travelled ???? during this time.
?
2?? sin ?? ?? ·?? =
?? 2
sin 2?? ?? -
?? 2
sin 2?? ?? ? sin 2?? =sin 2?? +
2?? ?? sin ??
2.2 A particle is projected with velocity ?? from the cusp of a smooth inverted
cyclcid down the arc. Show that the time of reaching the vertex is
?? v
?? ?? ?????? -?? (
?? ?? v????
)
where ?? is the radius of the generating circle.
(2009 : 10 Marks)
Solution:
Let particle se projected with velocity ?? from cusp ?? .
At any point ?? the equation of motion is
?? ?? 2
?? ?? ?? 2
=-???? sin ?? (??)
?? ?? 2
?? =?? -???? cos ?? (???? )
where ?? =???? is the arc length.
Using ?? =4?? sin ?? (equation of cycloid) (i) becomes
?? ?? 2
?? ?? ?? 2
=-????
?? 4?? ?
?? 2
?? ?? ?? 2
=-
?? 4?? ??
Multiplying by 2
????
????
and integrating
(
????
????
)
2
=
-?? 4?? ?? 2
+??
Page 3
Edurev123
2. Motion in a Pane
2.1 A shot fired with a velocity ?? at an angle (elevation) ?? strikes a point ?? in a
horizontal plane through the point of projection. If the point ?? is receding from the
gun with velocity ?? show that the elevation must be changed to ?? where
?????? ?? ?? =?????? ?? ?? +
?? ?? ?? ?????? ??
(2009: 12 Marks)
Solution:
Let ?? be the original position of the point.
When ?? is at rest,
???? =
?? 2
sin 2?? ??
Let the elevation be changed to ?? when the point recedes to hit it and let the point be hit
at ?? .
Then
???? =
?? 2
sin 2?? ?? ?? h?? ???????? ???????????? ???? ???????????? =
?? 2
sin 2?? ?? ·?? cos ?? =
2?? sin ?? ??
? So the point must have travelled ???? during this time.
?
2?? sin ?? ?? ·?? =
?? 2
sin 2?? ?? -
?? 2
sin 2?? ?? ? sin 2?? =sin 2?? +
2?? ?? sin ??
2.2 A particle is projected with velocity ?? from the cusp of a smooth inverted
cyclcid down the arc. Show that the time of reaching the vertex is
?? v
?? ?? ?????? -?? (
?? ?? v????
)
where ?? is the radius of the generating circle.
(2009 : 10 Marks)
Solution:
Let particle se projected with velocity ?? from cusp ?? .
At any point ?? the equation of motion is
?? ?? 2
?? ?? ?? 2
=-???? sin ?? (??)
?? ?? 2
?? =?? -???? cos ?? (???? )
where ?? =???? is the arc length.
Using ?? =4?? sin ?? (equation of cycloid) (i) becomes
?? ?? 2
?? ?? ?? 2
=-????
?? 4?? ?
?? 2
?? ?? ?? 2
=-
?? 4?? ??
Multiplying by 2
????
????
and integrating
(
????
????
)
2
=
-?? 4?? ?? 2
+??
At ?? ,?? =4?? and
????
????
=??
??? 2
=-4???? +?? ??? =?? 2
+4????
? (
????
????
)
2
=?? 2
+4???? -
?? 4?? ?? 2
????
????
=-v
?? 4?? [
4?? ?? (?? 2
+4???? )-?? 2
]
1/2
Taking -vi sign as body moves in direction of decreasing ?? .
???? =-2v
?? ?? ????
v
4?? ?? (?? 2
+4???? )-?? 2
??
?? 0
????? =-2v
?? ?? ? ?
?? =0
?? =
?? ?? ?
????
v
4?? ?? (?? 2
+4???? )-?? 2
?? 1
=2v
?? ?? [
sin
-1
?? v
4?? ?? (?? 2
+4???? )
]
0
4?? =2v
?? ?? sin
-1
v4????
v?? 2
+4????
=2v
?? ?? cot
-1
v?? 2
v4????
=2v
?? ?? cot
-1
?? 2v????
2.3. On a rigid body, the forces ???? (??ˆ+?? ??ˆ+?? ??ˆ
)?? ,?? (-?? ??ˆ-??ˆ+?? ??ˆ
)?? and ?? (?? ??ˆ+?? ??ˆ-
??ˆ
)?? are acting at pcints with position vector ??ˆ-??ˆ,?? ??ˆ+?? ??ˆ
and ?? ??ˆ-??ˆ
respectively.
Reduce this system to a single force ????
acting at the point (?? ??ˆ+?? ??ˆ) together with a
couple ????
whose axis passes through this point. Does the point (?? ??ˆ+?? ??ˆ) lie on the
central axis.
(2009 : 15 Marks)
Solution:
Let ?? (??ˆ-??ˆ),?? (2??ˆ+5??ˆ
) and ?? (4??ˆ-??ˆ
) be the given points and ?? 4??ˆ+2??ˆ) the point about
which the resultant has to found.
Then,
Page 4
Edurev123
2. Motion in a Pane
2.1 A shot fired with a velocity ?? at an angle (elevation) ?? strikes a point ?? in a
horizontal plane through the point of projection. If the point ?? is receding from the
gun with velocity ?? show that the elevation must be changed to ?? where
?????? ?? ?? =?????? ?? ?? +
?? ?? ?? ?????? ??
(2009: 12 Marks)
Solution:
Let ?? be the original position of the point.
When ?? is at rest,
???? =
?? 2
sin 2?? ??
Let the elevation be changed to ?? when the point recedes to hit it and let the point be hit
at ?? .
Then
???? =
?? 2
sin 2?? ?? ?? h?? ???????? ???????????? ???? ???????????? =
?? 2
sin 2?? ?? ·?? cos ?? =
2?? sin ?? ??
? So the point must have travelled ???? during this time.
?
2?? sin ?? ?? ·?? =
?? 2
sin 2?? ?? -
?? 2
sin 2?? ?? ? sin 2?? =sin 2?? +
2?? ?? sin ??
2.2 A particle is projected with velocity ?? from the cusp of a smooth inverted
cyclcid down the arc. Show that the time of reaching the vertex is
?? v
?? ?? ?????? -?? (
?? ?? v????
)
where ?? is the radius of the generating circle.
(2009 : 10 Marks)
Solution:
Let particle se projected with velocity ?? from cusp ?? .
At any point ?? the equation of motion is
?? ?? 2
?? ?? ?? 2
=-???? sin ?? (??)
?? ?? 2
?? =?? -???? cos ?? (???? )
where ?? =???? is the arc length.
Using ?? =4?? sin ?? (equation of cycloid) (i) becomes
?? ?? 2
?? ?? ?? 2
=-????
?? 4?? ?
?? 2
?? ?? ?? 2
=-
?? 4?? ??
Multiplying by 2
????
????
and integrating
(
????
????
)
2
=
-?? 4?? ?? 2
+??
At ?? ,?? =4?? and
????
????
=??
??? 2
=-4???? +?? ??? =?? 2
+4????
? (
????
????
)
2
=?? 2
+4???? -
?? 4?? ?? 2
????
????
=-v
?? 4?? [
4?? ?? (?? 2
+4???? )-?? 2
]
1/2
Taking -vi sign as body moves in direction of decreasing ?? .
???? =-2v
?? ?? ????
v
4?? ?? (?? 2
+4???? )-?? 2
??
?? 0
????? =-2v
?? ?? ? ?
?? =0
?? =
?? ?? ?
????
v
4?? ?? (?? 2
+4???? )-?? 2
?? 1
=2v
?? ?? [
sin
-1
?? v
4?? ?? (?? 2
+4???? )
]
0
4?? =2v
?? ?? sin
-1
v4????
v?? 2
+4????
=2v
?? ?? cot
-1
v?? 2
v4????
=2v
?? ?? cot
-1
?? 2v????
2.3. On a rigid body, the forces ???? (??ˆ+?? ??ˆ+?? ??ˆ
)?? ,?? (-?? ??ˆ-??ˆ+?? ??ˆ
)?? and ?? (?? ??ˆ+?? ??ˆ-
??ˆ
)?? are acting at pcints with position vector ??ˆ-??ˆ,?? ??ˆ+?? ??ˆ
and ?? ??ˆ-??ˆ
respectively.
Reduce this system to a single force ????
acting at the point (?? ??ˆ+?? ??ˆ) together with a
couple ????
whose axis passes through this point. Does the point (?? ??ˆ+?? ??ˆ) lie on the
central axis.
(2009 : 15 Marks)
Solution:
Let ?? (??ˆ-??ˆ),?? (2??ˆ+5??ˆ
) and ?? (4??ˆ-??ˆ
) be the given points and ?? 4??ˆ+2??ˆ) the point about
which the resultant has to found.
Then,
Resultant Force =S??
=10(??ˆ+2??ˆ+2??ˆ
)+5(-2??ˆ-??ˆ-2??ˆ
)+6(2??ˆ+2??ˆ-??ˆ
)
=12??ˆ-27??ˆ+24??ˆ
Positions vectors of ?? ,?? and ?? about ?? .
????
?????
=(??ˆ-??ˆ)-(4??ˆ+2??ˆ)=-3??ˆ-3??ˆ
????
?????
=(2??ˆ+5??ˆ
)-(4??ˆ+2??ˆ)=-2??ˆ-2??ˆ+5??ˆ
????
?????
=(4??ˆ-??ˆ
)-(4??ˆ+2??ˆ)=-2??ˆ-??ˆ
? Resuitant Couple =????
?????
×??
1
+????
?????
×??
2
+????
?????
×??
3
????
?????
×??
1
=|
??ˆ ??ˆ ??ˆ
-3 -3 0
10 20 20
|=-60??ˆ+60??ˆ-30??ˆ
????
?????
×??
2
=|
??ˆ ??ˆ ??ˆ
-2 -2 5
-10 -5 10
|=5??ˆ-30??ˆ-10??ˆ
????
?????
×??
3
=|
??ˆ ??ˆ ??ˆ
0 -2 -1
12 12 -6
|=-12??ˆ+24??ˆ
??
=-55??ˆ+18??ˆ-16??ˆ
is the resultant moment.
The resultant couple is not zero so the point does not lie on the control axis.
2.4 If ?? ?? ,?? ?? ,?? ?? are the velocities at three points ?? ,?? , Cof the parn of a projective,
where the inclinations to the horizon are ?? ,?? -?? ,?? -???? and if ?? ?? ,???
?? are the times
describing the arcs ???? ,???? respectively, prove that ?? ?? ?? ?? =?? ?? ?? ?? and
?? ?? ?? +
?? ?? ?? =
???????? ?? ?? ??
(2010 : 12 Marks)
Solution:
Above figure shows the path of projectile given in question.
As in ?? -direction, there is no acceleration.
? Velocities in ?? -direction at ?? ,?? ,?? are equal.
Page 5
Edurev123
2. Motion in a Pane
2.1 A shot fired with a velocity ?? at an angle (elevation) ?? strikes a point ?? in a
horizontal plane through the point of projection. If the point ?? is receding from the
gun with velocity ?? show that the elevation must be changed to ?? where
?????? ?? ?? =?????? ?? ?? +
?? ?? ?? ?????? ??
(2009: 12 Marks)
Solution:
Let ?? be the original position of the point.
When ?? is at rest,
???? =
?? 2
sin 2?? ??
Let the elevation be changed to ?? when the point recedes to hit it and let the point be hit
at ?? .
Then
???? =
?? 2
sin 2?? ?? ?? h?? ???????? ???????????? ???? ???????????? =
?? 2
sin 2?? ?? ·?? cos ?? =
2?? sin ?? ??
? So the point must have travelled ???? during this time.
?
2?? sin ?? ?? ·?? =
?? 2
sin 2?? ?? -
?? 2
sin 2?? ?? ? sin 2?? =sin 2?? +
2?? ?? sin ??
2.2 A particle is projected with velocity ?? from the cusp of a smooth inverted
cyclcid down the arc. Show that the time of reaching the vertex is
?? v
?? ?? ?????? -?? (
?? ?? v????
)
where ?? is the radius of the generating circle.
(2009 : 10 Marks)
Solution:
Let particle se projected with velocity ?? from cusp ?? .
At any point ?? the equation of motion is
?? ?? 2
?? ?? ?? 2
=-???? sin ?? (??)
?? ?? 2
?? =?? -???? cos ?? (???? )
where ?? =???? is the arc length.
Using ?? =4?? sin ?? (equation of cycloid) (i) becomes
?? ?? 2
?? ?? ?? 2
=-????
?? 4?? ?
?? 2
?? ?? ?? 2
=-
?? 4?? ??
Multiplying by 2
????
????
and integrating
(
????
????
)
2
=
-?? 4?? ?? 2
+??
At ?? ,?? =4?? and
????
????
=??
??? 2
=-4???? +?? ??? =?? 2
+4????
? (
????
????
)
2
=?? 2
+4???? -
?? 4?? ?? 2
????
????
=-v
?? 4?? [
4?? ?? (?? 2
+4???? )-?? 2
]
1/2
Taking -vi sign as body moves in direction of decreasing ?? .
???? =-2v
?? ?? ????
v
4?? ?? (?? 2
+4???? )-?? 2
??
?? 0
????? =-2v
?? ?? ? ?
?? =0
?? =
?? ?? ?
????
v
4?? ?? (?? 2
+4???? )-?? 2
?? 1
=2v
?? ?? [
sin
-1
?? v
4?? ?? (?? 2
+4???? )
]
0
4?? =2v
?? ?? sin
-1
v4????
v?? 2
+4????
=2v
?? ?? cot
-1
v?? 2
v4????
=2v
?? ?? cot
-1
?? 2v????
2.3. On a rigid body, the forces ???? (??ˆ+?? ??ˆ+?? ??ˆ
)?? ,?? (-?? ??ˆ-??ˆ+?? ??ˆ
)?? and ?? (?? ??ˆ+?? ??ˆ-
??ˆ
)?? are acting at pcints with position vector ??ˆ-??ˆ,?? ??ˆ+?? ??ˆ
and ?? ??ˆ-??ˆ
respectively.
Reduce this system to a single force ????
acting at the point (?? ??ˆ+?? ??ˆ) together with a
couple ????
whose axis passes through this point. Does the point (?? ??ˆ+?? ??ˆ) lie on the
central axis.
(2009 : 15 Marks)
Solution:
Let ?? (??ˆ-??ˆ),?? (2??ˆ+5??ˆ
) and ?? (4??ˆ-??ˆ
) be the given points and ?? 4??ˆ+2??ˆ) the point about
which the resultant has to found.
Then,
Resultant Force =S??
=10(??ˆ+2??ˆ+2??ˆ
)+5(-2??ˆ-??ˆ-2??ˆ
)+6(2??ˆ+2??ˆ-??ˆ
)
=12??ˆ-27??ˆ+24??ˆ
Positions vectors of ?? ,?? and ?? about ?? .
????
?????
=(??ˆ-??ˆ)-(4??ˆ+2??ˆ)=-3??ˆ-3??ˆ
????
?????
=(2??ˆ+5??ˆ
)-(4??ˆ+2??ˆ)=-2??ˆ-2??ˆ+5??ˆ
????
?????
=(4??ˆ-??ˆ
)-(4??ˆ+2??ˆ)=-2??ˆ-??ˆ
? Resuitant Couple =????
?????
×??
1
+????
?????
×??
2
+????
?????
×??
3
????
?????
×??
1
=|
??ˆ ??ˆ ??ˆ
-3 -3 0
10 20 20
|=-60??ˆ+60??ˆ-30??ˆ
????
?????
×??
2
=|
??ˆ ??ˆ ??ˆ
-2 -2 5
-10 -5 10
|=5??ˆ-30??ˆ-10??ˆ
????
?????
×??
3
=|
??ˆ ??ˆ ??ˆ
0 -2 -1
12 12 -6
|=-12??ˆ+24??ˆ
??
=-55??ˆ+18??ˆ-16??ˆ
is the resultant moment.
The resultant couple is not zero so the point does not lie on the control axis.
2.4 If ?? ?? ,?? ?? ,?? ?? are the velocities at three points ?? ,?? , Cof the parn of a projective,
where the inclinations to the horizon are ?? ,?? -?? ,?? -???? and if ?? ?? ,???
?? are the times
describing the arcs ???? ,???? respectively, prove that ?? ?? ?? ?? =?? ?? ?? ?? and
?? ?? ?? +
?? ?? ?? =
???????? ?? ?? ??
(2010 : 12 Marks)
Solution:
Above figure shows the path of projectile given in question.
As in ?? -direction, there is no acceleration.
? Velocities in ?? -direction at ?? ,?? ,?? are equal.
?? 1
cos?? =?? 2
cos(?? -?? )=?? 3
cos(?? -2?? ) (1)
Now,
1
?? 1
+
1
?? 3
=
cos ?? ?? 2
cos (?? -?? )
+
cos (?? -2?? )
?? 2
cos (?? -?? )
=
cos ?? +cos (?? -2?? )
?? 2
cos (?? -?? )
=
2cos (?? -?? )·cos ?? ?? 2
cos (?? -?? )
=
2cos ?? ?? 2
1
?? 1
+
1
?? 3
=
2cos ?? ?? 2
, which is the required result
Now, Newton's equation of motion is ?? -direction with ?? as acceleration.
and
?? 3
sin (?? -2?? )=?? 2
sin ?? -?? )-?? ?? 2
for ????
?? 2
sin (?? -?? )=?? 1
sin ?? -?? ?? 1
for ????
??? 1
=
?? 1
sin ?? -?? 2
sin (?? -?? )
??
and
?? 2
=
?? 2
sin (?? -?? )-?? 3
sin (?? -2?? )
??
So,
?? 3
?? 1
=
?? 3
?? 1
sin ?? -?? 3
?? 2
sin (?? -?? )
?? (3)
=
?? 1
?? 2
cos (?? -?? )sin ?? cos (?? -2?? )
-
?? 1
?? 2
cos ?? sin (?? -?? )
cos (?? -2?? )
??
(from (1))
=
?? 1
V
2
?? cos (?? -2?? )
{sin ?? cos (?? -?? )-cos ?? sin (?? -?? )}
=
?? 1
?? 2
sin ?? ?? cos (?? -2?? )
=LHS
Also,
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Dec 22, 2024 Last updated |
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