Question

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless surface with a tangential speed of 5 m/s. The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to 0.250 m. What is the instantaneous speed of the mass at the moment the string reaches a length of 0.250 m?

Answer 1

15 m/s

Step-by-step explanation:

L = mvr

Li = (2.00 kg)(0.750 m)(5m/s) = 7.5 kgm^2/s

conservation of angular momentum --> Li=Lf

Lf = 7.5 kgm^2/s

7.5 kgm^2/s = (2.00 kg)(0.250 m)(vf)

vf = 15 m/s

Answer 2

`v_f = 15 (m)/(s)`

Step-by-step explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum
`\vec{L}` is

`\vec{L}  = \vec{r} * \vec{p}`

where
`\vec{r}` is the position and
`\vec{p}` the linear momentum.

We also know that the torque is

`\vec{\tau} = \frac{d\vec{L}}{dt}  = (d)/(dt) ( \vec{r} * \vec{p} )`

`\vec{\tau} =  (d)/(dt)  \vec{r} * \vec{p} +   \vec{r} * (d)/(dt) \vec{p}`

`\vec{\tau} =  \vec{v} * \vec{p} +   \vec{r} * \vec{F}`

but, as the linear momentum is
`\vec{p} = m \vec{v}` this means that is parallel to the velocity, and the first term must equal zero

`\vec{v} * \vec{p}=0`

so

`\vec{\tau} =   \vec{r} * \vec{F}`

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

`\vec{\tau}_(rod) =   0`

this means, for the angular momentum measure from the rod:

`\frac{d\vec{L}_(rod)}{dt} =   0`

that means :

`\vec{L}_(rod) = constant`

So, the magnitude of initial angular momentum is :

`| \vec{L}_(rod_i) | = |\vec{r}_i||\vec{p}_i| cos(\theta)`

but the angle is 90°, so:

`| \vec{L}_(rod_i) | = |\vec{r}_i||\vec{p}_i|`

`| \vec{L}_(rod_i) | = r_i * m * v_i`

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

`| \vec{L}_(rod_i) | = 0.750 \ m \ 2.00 \ kg \ 5 \ (m)/(s)`

`| \vec{L}_(rod_i) | = 7.5 (kg m^2)/(s)`

For our final angular momentum we have:

`| \vec{L}_(rod_f) | = r_f * m * v_f`

and the radius is 0.250 m and the mass is 2.00 kg

`| \vec{L}_(rod_f) | = 0.250 m * 2.00 kg * v_f`

but, as the angular momentum is constant, this must be equal to the initial angular momentum

`7.5 (kg m^2)/(s) = 0.250 m * 2.00 kg * v_f`

`v_f = (7.5 (kg m^2)/(s))/( 0.250 m * 2.00 kg)`

`v_f = 15 (m)/(s)`