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?

1 Answer
Aug 13, 2017

Answer:

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

Explanation:

Diagram:

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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)