Why does flipping a pencil across a table make it move like this?
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I've always thought it was weird that pencils act like this: if one pulls their finger along the side of a pencil until it touches the surface below, the pencil is launched in the opposite direction of the way that the finger moved. Why is this?
kinematics
asked 3 hours ago
StormblessedStormblessed
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add a comment |
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I've always thought it was weird that pencils act like this: if one pulls their finger along the side of a pencil until it touches the surface below, the pencil is launched in the opposite direction of the way that the finger moved. Why is this?
kinematics
asked 3 hours ago
StormblessedStormblessed
1214
New contributor
Stormblessed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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may be -> linear momentum dies, still rotational momentum is there and hence 'launched in opposite direction' :)
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– aranyak
3 hours ago
add a comment |
$begingroup$
I've always thought it was weird that pencils act like this: if one pulls their finger along the side of a pencil until it touches the surface below, the pencil is launched in the opposite direction of the way that the finger moved. Why is this?
kinematics
asked 3 hours ago
StormblessedStormblessed
1214
New contributor
Stormblessed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I've always thought it was weird that pencils act like this: if one pulls their finger along the side of a pencil until it touches the surface below, the pencil is launched in the opposite direction of the way that the finger moved. Why is this?
kinematics
kinematics
asked 3 hours ago
StormblessedStormblessed
1214
New contributor
Stormblessed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
StormblessedStormblessed
1214
New contributor
Stormblessed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
StormblessedStormblessed
1214
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asked 3 hours ago
StormblessedStormblessed
1214
asked 3 hours ago
StormblessedStormblessed
1214
1214
New contributor
Stormblessed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Stormblessed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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may be -> linear momentum dies, still rotational momentum is there and hence 'launched in opposite direction' :)
$endgroup$
– aranyak
3 hours ago
add a comment |
$begingroup$
may be -> linear momentum dies, still rotational momentum is there and hence 'launched in opposite direction' :)
$endgroup$
– aranyak
3 hours ago
$begingroup$
may be -> linear momentum dies, still rotational momentum is there and hence 'launched in opposite direction' :)
$endgroup$
– aranyak
3 hours ago
$begingroup$
may be -> linear momentum dies, still rotational momentum is there and hence 'launched in opposite direction' :)
$endgroup$
– aranyak
3 hours ago
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
As the finger comes down the side of the pencil, two things happen:
A compression that imparts a horizontal force taking the pencil away from the finger
And a rotation that causes the pencil to rotate tending to bring the pencil back to the start point
These combine to define how far the pencil travels before it stops.
The use of spin can be seen on a snooker or billiards table : top, bottom and side...
answered 3 hours ago
Solar MikeSolar Mike
1313
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Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
The sequence of events is shown below.
Initially the pencil is propelled forward with speed $v_{rm A}$ but has backspin $omega_{rm A}$ (anticlockwise rotation) so there is relative movement between the pencil and the surface as $v_{rm A} ne romega_{rm A}$ where $r$ is the radius of the pencil.
A kinetic friction force acts which reduces the rotational speed of the pencil $omega_{rm B}$ until there is no rotation of the pencil $omega_{rm C}=0$ but the pencil is still moving forward $v_{rm C}$.
The frictional force then starts the pencil rotating clockwise with increasing angular speed and eventually the no slip condition, $v_{rm D} = romega_{rm D}$, is reached.
answered 43 mins ago
FarcherFarcher
52.5k340112
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As the finger comes down the side of the pencil, two things happen:
A compression that imparts a horizontal force taking the pencil away from the finger
And a rotation that causes the pencil to rotate tending to bring the pencil back to the start point
These combine to define how far the pencil travels before it stops.
The use of spin can be seen on a snooker or billiards table : top, bottom and side...
answered 3 hours ago
Solar MikeSolar Mike
1313
New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As the finger comes down the side of the pencil, two things happen:
A compression that imparts a horizontal force taking the pencil away from the finger
And a rotation that causes the pencil to rotate tending to bring the pencil back to the start point
These combine to define how far the pencil travels before it stops.
The use of spin can be seen on a snooker or billiards table : top, bottom and side...
answered 3 hours ago
Solar MikeSolar Mike
1313
New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
As the finger comes down the side of the pencil, two things happen:
A compression that imparts a horizontal force taking the pencil away from the finger
And a rotation that causes the pencil to rotate tending to bring the pencil back to the start point
These combine to define how far the pencil travels before it stops.
The use of spin can be seen on a snooker or billiards table : top, bottom and side...
answered 3 hours ago
Solar MikeSolar Mike
1313
New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As the finger comes down the side of the pencil, two things happen:
A compression that imparts a horizontal force taking the pencil away from the finger
And a rotation that causes the pencil to rotate tending to bring the pencil back to the start point
These combine to define how far the pencil travels before it stops.
The use of spin can be seen on a snooker or billiards table : top, bottom and side...
answered 3 hours ago
Solar MikeSolar Mike
1313
New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Solar MikeSolar Mike
1313
New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Solar MikeSolar Mike
1313
answered 3 hours ago
Solar MikeSolar Mike
1313
1313
New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Solar Mike is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
add a comment |
$begingroup$
The sequence of events is shown below.
Initially the pencil is propelled forward with speed $v_{rm A}$ but has backspin $omega_{rm A}$ (anticlockwise rotation) so there is relative movement between the pencil and the surface as $v_{rm A} ne romega_{rm A}$ where $r$ is the radius of the pencil.
A kinetic friction force acts which reduces the rotational speed of the pencil $omega_{rm B}$ until there is no rotation of the pencil $omega_{rm C}=0$ but the pencil is still moving forward $v_{rm C}$.
The frictional force then starts the pencil rotating clockwise with increasing angular speed and eventually the no slip condition, $v_{rm D} = romega_{rm D}$, is reached.
answered 43 mins ago
FarcherFarcher
52.5k340112
$endgroup$
add a comment |
$begingroup$
The sequence of events is shown below.
Initially the pencil is propelled forward with speed $v_{rm A}$ but has backspin $omega_{rm A}$ (anticlockwise rotation) so there is relative movement between the pencil and the surface as $v_{rm A} ne romega_{rm A}$ where $r$ is the radius of the pencil.
A kinetic friction force acts which reduces the rotational speed of the pencil $omega_{rm B}$ until there is no rotation of the pencil $omega_{rm C}=0$ but the pencil is still moving forward $v_{rm C}$.
The frictional force then starts the pencil rotating clockwise with increasing angular speed and eventually the no slip condition, $v_{rm D} = romega_{rm D}$, is reached.
answered 43 mins ago
FarcherFarcher
52.5k340112
$endgroup$
add a comment |
$begingroup$
The sequence of events is shown below.
Initially the pencil is propelled forward with speed $v_{rm A}$ but has backspin $omega_{rm A}$ (anticlockwise rotation) so there is relative movement between the pencil and the surface as $v_{rm A} ne romega_{rm A}$ where $r$ is the radius of the pencil.
A kinetic friction force acts which reduces the rotational speed of the pencil $omega_{rm B}$ until there is no rotation of the pencil $omega_{rm C}=0$ but the pencil is still moving forward $v_{rm C}$.
The frictional force then starts the pencil rotating clockwise with increasing angular speed and eventually the no slip condition, $v_{rm D} = romega_{rm D}$, is reached.
answered 43 mins ago
FarcherFarcher
52.5k340112
$endgroup$
The sequence of events is shown below.
Initially the pencil is propelled forward with speed $v_{rm A}$ but has backspin $omega_{rm A}$ (anticlockwise rotation) so there is relative movement between the pencil and the surface as $v_{rm A} ne romega_{rm A}$ where $r$ is the radius of the pencil.
A kinetic friction force acts which reduces the rotational speed of the pencil $omega_{rm B}$ until there is no rotation of the pencil $omega_{rm C}=0$ but the pencil is still moving forward $v_{rm C}$.
The frictional force then starts the pencil rotating clockwise with increasing angular speed and eventually the no slip condition, $v_{rm D} = romega_{rm D}$, is reached.
answered 43 mins ago
FarcherFarcher
52.5k340112
answered 43 mins ago
FarcherFarcher
52.5k340112
answered 43 mins ago
FarcherFarcher
52.5k340112
answered 43 mins ago
FarcherFarcher
52.5k340112
52.5k340112
add a comment |
add a comment |
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$begingroup$
may be -> linear momentum dies, still rotational momentum is there and hence 'launched in opposite direction' :)
$endgroup$
– aranyak
3 hours ago