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
Answer:
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=TF_G=ma_c#
We also have that:
#a_c=v^2/r#
Since we know that
#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.
#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
Since the height
#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
#=>=(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)