Total linear momentum of swinging bat

[Juke]

Homework Statement

A bat for use in a ball game has total mass M = 1:5 kg and total length ` = 1 m;
assume it can be approximated by a thin uniform rod of length `. What is its radius
of gyration k[itex]_{0}[/itex] about an axis perpendicular to the bat and through its centre of
mass? A player grips the bat and swings it about an axis 0.8m above the top of the
handle and perpendicular to the bat (see diagram); what is its moment of inertia
about the rotation axis?

While swinging the bat, the player strikes a ball with the bat vertical (see diagram).
If the bottom end of the bat is moving horizontally at 10 ms1 at the moment of
impact,find at that instant (i) the angular velocity of the bat, (ii) the angular
momentum of the bat about the rotation axis, (iii) the bat's kinetic energy and (iv)
its total linear momentum.

The player has been told by a coach to aim to strike the ball in such a way that it
needs no impulse from his hands at the moment of collision. Suppose the ball strikes
the bat a distance d below the centre of mass (see diagram). Assuming that the
impulse J from the ball is perpendicular to the bat, find in terms of J the changes
during the collision in both the angular velocity of the bat and the linear velocity
of its centre of mass. If the player grips the bat a distance d[itex]_{grip}[/itex] above the centre
of mass, show that the point of grip suffers no sudden change in velocity only if
d = k[itex]^{2}_{0}[/itex]/d[itex]_{grip}[/itex]
. If the player grips the bat 0.1m below the top, use this result to find
the value of d he should aim for.

Homework Equations

p =mv
L = Iω
K =1/2 * Iω[itex]^{2}[/itex]
v= ωr
I = mk[itex]^{2}[/itex]
I = I[itex]_{0}[/itex] + Md[itex]^{2}[/itex]

The Attempt at a Solution

I'm stuck on part (iv), I don't know how to find the total linear momentum.

What I have done so far is tried using p=mv, with v = 10 but it gave a wrong answer. How can a rotating object have linear momentum?

The answers p = 10.83 kgms[itex]^{-1}[/itex]

 

[SammyS]

Homework Statement

A bat for use in a ball game has total mass M = 1:5 kg and total length ` = 1 m;
assume it can be approximated by a thin uniform rod of length `. What is its radius
of gyration k[itex]_{0}[/itex] about an axis perpendicular to the bat and through its centre of
mass? A player grips the bat and swings it about an axis 0.8m above the top of the
handle and perpendicular to the bat (see diagram); what is its moment of inertia
about the rotation axis?

While swinging the bat, the player strikes a ball with the bat vertical (see diagram).
If the bottom end of the bat is moving horizontally at 10 ms1 at the moment of
impact,find at that instant (i) the angular velocity of the bat, (ii) the angular
momentum of the bat about the rotation axis, (iii) the bat's kinetic energy and (iv)
its total linear momentum.

The player has been told by a coach to aim to strike the ball in such a way that it
needs no impulse from his hands at the moment of collision. Suppose the ball strikes
the bat a distance d below the centre of mass (see diagram). Assuming that the
impulse J from the ball is perpendicular to the bat, find in terms of J the changes
during the collision in both the angular velocity of the bat and the linear velocity
of its centre of mass. If the player grips the bat a distance d[itex]_{grip}[/itex] above the centre
of mass, show that the point of grip suffers no sudden change in velocity only if
d = k[itex]^{2}_{0}[/itex]/d[itex]_{grip}[/itex]
. If the player grips the bat 0.1m below the top, use this result to find
the value of d he should aim for.

Homework Equations

p =mv
L = Iω
K =1/2 * Iω[itex]^{2}[/itex]
v= ωr
I = mk[itex]^{2}[/itex]
I = I[itex]_{0}[/itex] + Md[itex]^{2}[/itex]

The Attempt at a Solution

I'm stuck on part (iv), I don't know how to find the total linear momentum.

What I have done so far is tried using p=mv, with v = 10 but it gave a wrong answer. How can a rotating object have linear momentum?

The answers p = 10.83 kgms[itex]^{-1}[/itex]

It might help if you would show your answers for the earlier questions.


Where is the center of mass for the bat?

What is the velocity of the center of mass of the bat?

 

[Juke]

It might help if you would show your answers for the earlier questions.


Where is the center of mass for the bat?

What is the velocity of the center of mass of the bat?

Radius of gyration about axis through centre of mass: 0.29m; moment of inertia about actual rotation axis 2.66 kg m2.
(a) Angular velocity 5.56 rad s−1;
(b) Angular momentum 14.78 kg m2 s−1;
(c) Kinetic energy 41.05 J;
(d) Total momentum 10.83 kg m s−1.

The centre of mass of the bat is at the mid point of the bat.

I don't know its velocity, never have I come across a question where you needed to find the Vcm for a single body.

 

[SammyS]

Radius of gyration about axis through centre of mass: 0.29m; moment of inertia about actual rotation axis 2.66 kg m2.
(a) Angular velocity 5.56 rad s−1;
(b) Angular momentum 14.78 kg m2 s−1;
(c) Kinetic energy 41.05 J;
(d) Total momentum 10.83 kg m s−1.

The centre of mass of the bat is at the mid point of the bat.

I don't know its velocity, never have I come across a question where you needed to find the Vcm for a single body.

So, how did you find the linear momentum?

 

[Juke]

So, how did you find the linear momentum?

I took a guess and tried using the horizontal velocity of 10m/s but that's incorrect as you can see from the answers.

So to find the total linear momentum I need the Vcm?

 

[SammyS]

I took a guess and tried using the horizontal velocity of 10m/s but that's incorrect as you can see from the answers.

So to find the total linear momentum I need the Vcm?

How far is the CM from the axis of rotation?

 

[Juke]

How far is the CM from the axis of rotation?

0.8+0.5 = 1.3m

 

[SammyS]

0.8+0.5 = 1.3m

Use that together with the angular velocity to find V_CM .

 

[Juke]

Use that together with the angular velocity to find V_CM .

Yes, got it now. Thanks

Vcm = 1.3 * 10/1.8

P = 1.5 * Vcm = 10.83 Ns