Calc Direction of Tension: Work & Kinetic Energy

[Elias Waranoi]

Homework Statement

A block with mass m is revolving with linear speed v₁ in a circle of radius r₁ on a frictionless horizontal surface. The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to r₂. (a) Calculate the tension T in the string as a function of r, the distance of the block from the hole. Your answer will be in terms of the initial velocity v₁ and the radius r₁. (b) Use W = ∫_r1^r₂T(r) ⋅ dr to calculate the work done by T(r) when r changes from r₁ to r₂. (c) Compare the results of part (b) to the change in the kinetic energy of the block

Homework Equations

3. Attempt at solution
I've solved the question except that I thought that T ⋅ dr = Tdr while it was supposed to T ⋅ dr = -Tdr because apparently T and dr are antiparallel. I thought that tension is only in the direction towards the middle of a rope and we're obviously pulling r closer to the middle so how come they are antiparallel?

 

[Doc Al]

Note that dr points in the direction of increasing r, which is outward.

 

[Elias Waranoi]

Oh. You mean dr is not just a change in length but specifically a positive change? So -dr would ALWAYS be a negative change?

 

[BvU]

$r_1$ goes to $r_2$ for the integral. Since $r_1<r_2$, that minus sign should come out of the integral.

 

[ehild]

3. Attempt at solution
I've solved the question except that I thought that T ⋅ dr = Tdr while it was supposed to T ⋅ dr = -Tdr because apparently T and dr are antiparallel. I thought that tension is only in the direction towards the middle of a rope and we're obviously pulling r closer to the middle so how come they are antiparallel?

$\vec r$ is the position vector, pointing to the block, and $\vec T$ is the force of tension which points towards the center. They are anti-parallel. $\vec {dr}$ is the change of ${\vec r }$. The work done by the tension is $W =\int_{r_1}^{r_2}{\vec T \cdot \vec {dr}}$ Written with the magnitudes T and dr, $W = \int_{r_1}^{r_2}{T dr cos(\theta)} = \int_{r_1}^{r_2}{(-T dr) }$

 

[BvU]

Ehild is of course correct: $\hat T$ and $\hat r$ have to point in the same direction for the dot product to make sense.

 

[Elias Waranoi]

Thanks ehild. I fully get it now!

 

[kuruman]

Another way to look at it

For line integrals I find it least confusing to write the two vectors in the integrand in terms of unit vectors, then take the dot product, and finally do the integral. In this case, for example, I would write $\vec{T} \cdot d\vec{r}=(-T)\hat{r} \cdot (dr)\hat{r} = -Tdr$. So then, $\int_{r_1}^{r_2}{\vec T \cdot d\vec{r}} =- \int_{r_1}^{r_2}{T~dr}$. Note that $d\vec{r}$ is always $+(dr) \hat{r}$ regardless of whether the line integral is taken radially out or radially in. In other words, don't worry whether the two vectors are parallel or antiparallel; the minus sign, if needed, will be taken care of by the limits of integration.