Question

A `1"kg"` stone is tied to a `0.5"m"` string and swung in a vertical circle. What is the tension in the string at the bottom of the loop?

Answer 1

`T~~59" N"` (toward center)

Explanation:

Diagram:

where `vecT` is the force of tension and `vecF_G` is the force of gravity.

We have the following information:

• `|-> m=1" kg"`

• `|->r=0.5" m"`

• `|->g=9.81" m"//"s"^2`

We can take inventory of the forces acting on the stone:

`F_("net")=sumF_c=T-F_G=ma_c`

We also have that:

`a_c=v^2/r`

`=>T-F_G=(mv^2)/r`

Since we know that `F_G=mg`, we can solve for tension as:

`color(blue)(T=(mv^2)/r+mg)`

We aren't given a velocity, so in order to get a numerical answer, we'll have to look to energy conservation.

Conservation of energy:

`DeltaE=DeltaK+DeltaU+E_("th")`

In this case, at the top of the loop, the stone has both gravitational potential energy and kinetic energy.

• `U_g=mgh`

• `K=1/2mv^2`

At the bottom of the loop, the stone has only kinetic energy `(h=0)`.

Since the height `h` is equal to the diameter of the circle, we have `h=2r`. Therefore, we have:

`color(blue)(1/2mv_"top"^2+mg(2r)=1/2mv_"bottom"^2)`

At the top of the loop, we will assume that the velocity is the critical velocity, the minimum velocity necessary to maintain circular motion, where:

`v_"critical"=sqrt(rg)`

`=>color(blue)(1/2mrg+mg(2r)=1/2mv_"bottom"^2)`

We can solve for `v_"bottom"^2`:

`=>v_"bottom"^2=(2(1/2mrg+2mrg))/m`

`=>=(mrg+4mrg)/m`

`=>=color(blue)(5rg)`

Putting this into our equation for tension:

`T=(mv^2)/r+mg`

`=>=(m*5rg)/r+mg`

`=>=5mg+mg`

`=>=color(blue)(6mg)`

`=>=6(1"kg")(9.81"m"//"s"^2)`

`=>=58.86" N"`

`=>color(blue)(T~~59" N")` (toward center)