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I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simplesimply applying conservation of angular momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.

I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simple conservation of momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.

I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simply applying conservation of angular momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.

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Qmechanic
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I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simple conservation of momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about

solving solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.

I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simple conservation of momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about

solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.

I'm given the following very simple problem:

There is an object of mass $m$ attached to a string (whose mass can be neglected) on a table, whereby the string goes through the hole in the middle of the table and the particle moves on this frictionless horizontal table in a circular motion. It's initial angular velocity is $\omega_0$ and the initial radius/distance to the hole is $r_0$. Now you pull the string s.t. the distance to the hole gets smaller and is now $r$. What is the angular velocity $\omega_1$ now?

By simple conservation of momentum one obtains $\omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)^2$. However, I also thought about solving this by using the fact that the magnitude of the tangential velocity $v$ doesn't change during this process and hence I obtained $$\omega_0 r_0 = \omega_1 r \iff \omega_1 = \omega_0 \left(\dfrac{r_0}{r}\right)$$ which is clearly wrong. Why is the assumption that the tangential velocity doesn't change wrong? Intuitively I would assume that since I'm exerting a force parallel to the string that this doesn't affect the tangential component of the objects velocity.

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Jake
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Angular velocity: simple mistake in thinking processwhy does decreasing the radius affect the tangential velocity?

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Jake
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