A girl is spinning a Dr. Octopus doll (400 g) around in a vertical circle. The doll goes around once every 0.700s on the string which is 0.800m long.
What is the speed of the doll in m/s?
What is the tension on the string at the top of the circle?
What is the tension on the string at the bottom of the circle?
How slow can she go and still keep Dr. Octopus in a circle?
Can you please include a free body diagram if possible?
2 Answers | Add Yours
Please look at the picture image attached.
There are two forces acting on a doll, the gravity force `mg` and the tension force `F_T.` The gravity force is always directed downwards. The tension force is directed downwards at the upper point and upwards at the lowest point.
For objects in circular motion the net force is always directed towards the center and is equal to
where `r` is the radius and `V_p` is the speed at the specific point. This speed isn't constant for vertical circular motion! So the answer to the first question is different for different positions of a doll and cannot be computed as the path length divided by the time of one revolution.
At the upper point, `mg` and `F_T` have the same direction downwards, and the net force also. So
`mg+F_(T)=m*V_(u)^2/r` and `F_(T)=m*(V_(u)^2/r-g).`
A doll is kept at the circle if some tension remains, i.e. `V_(u)^2/rgt=g,`
`V_(u)^2gt=sqrt(rg) =2.8(m/s).` This is the answer to the last question.
At the lowest point `mg` and `F_T` have opposite directions, and
`F_(T)-mg=m*V_(l)^2/r,` or `F_(T)=m*(V_(l)^2/r+g).`
Now let's try to find `V_u` and `V_l.` They are necessary to answer questions 1, 2 and 3.
We'll use the Law of conservation of energy. There are two types of energy, kinetic and potential. Assume zero level of potential energy to be at the lowest point. Then
For the second equation you will have to use the given period of revolution.
how can substitute the values? of questions 1,2,3?? can you do the math again including the numbers and values
where are the numbers for the other questions ?
We’ve answered 319,641 questions. We can answer yours, too.Ask a question