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3 Answers
Cesareo R.
Aug 14, 2017

See below.

Explanation:

The equilibrium before the string break is attained at radius of rotation regarding the vertical of

#r = l sin theta#

with #l = 1.5# and #theta = 53^@# so the tangential velocity is

#v=omega r#

We know that considering the momenta regarding #A#

#l sin theta m g - m omega^2l cos theta= 0# then

#omega = sqrt(g tan theta)# then

#v = l sin theta sqrt(g tan theta) = 4.32# [m/s] (i)

Considering the movement referential at the bob, after the break the movement equations are

#(x,y) = (v t, -1/2 g t^2)# so the ground is hit at

#y = -(h-l cos theta) = -1/2 g t_0^2 # with #h = 3# which gives

#t_0 = 0.65# [s] (ii)

and the flying distance ground projected is

#v t_0 = 4.32 xx 0.65 = 2.82# [m] (iii)

P dilip_k
Aug 14, 2017

enter image source here

Given that

  • the length of the string in the conical pendulum #color(blue)(l=1.5m)#

  • the mass of the bob #color(blue)(m=2.5kg)#

  • the angle of inclination of the string during uniform circular motion of the bob is #color(blue)(theta=53^@)#

  • the height of point of suspension (#A#) of the bob #color(blue)(AC=3m)#

  • the radius of horizontal circular path #color(blue)(r=lsintheta=1.5sin53^@=1.5xx0.8=1.2m)#

  • #T# be the tension on the string

and

  • #v# be the uniform speed of the bob.

Here the bob will be in a dynamic equilibrium.

  • Horizontal component of the tension provides the the centripetal force required for uniform circular motion with

speed #v#

Hence #(mv^2)/r=Tsintheta....[1]#

  • The vertical component of the tension T will balance the weight of the bob

Hence #mg=Tcostheta.....[2]#

Now dividing [1] by [2] we get

#v^2/(rg)=sintheta/costheta#

#=>v^2/(1.2xx10)=0.8/0.6=4/3#

#=>v^2=4/3xx12=4^2#

#=>v=4m"/"s#

  • (i) So the tangential speed of the bob at the time of breaking of string is #color(red)(v=4m"/"s)#

  • Now the height of the horizontal circular path from the ground
    #h=AC-lcostheta#

#=>h=3-1.5xxcos53^@=3-0.9=2.1m#

If #t# be the time taken by the bob to reach at the ground after breaking of string ,then

#h=vxxsin0^@xxt+1/2xxgxxt^2# (since initial vertical component of velicity of projection is #vsin0^@=0#)

Hence #2.1=1/2xx10xxt^2#

#=>t=sqrt(2.1/5)~~0.65s#

(ii) So the required time #t=0.65s#

If #d# be the distance teversed by the bob with the horizontal tangential speed,then
#d=vxxt=4xx0.65=2.6m#

(iii) If the distance of the bob on the ground from the rod be #D#, then

#D=sqrt(d^2+r^2)#

#=>D=sqrt(2.6^2+1.2^2)~~2.86m#

A08
Aug 14, 2017

(i)
www.cpp.edu
The problem is depicted as a conical pendulum in the figure above. #theta=53^@# is the angle string of length #=1.5m# makes with the vertical. The tension in the string can be resolved into two components. The vertical component balances the weight of the bob #=2.5kg# as there is no acceleration in this direction.

#Tcostheta=mg# ......(1)

The horizontal component provides the necessary centripetal force.

#Tsin theta=(mv^2)/r# .....(2)
where #v# is the constant velocity of the bob.

From the figure Radius #r# of horizontal circle traced by bob at the time of breaking of string

#r=Lsintheta=1.5xxsin53^@#

Dividing (2) with (1)

#sintheta/costheta=v^2/(rg)#

Substituting value of #r# we get

#=>v^2/((1.5xxsin53^@)xx9.81)=(sin53^@)/(cos53^@)#

#=>v^2=(1.5xxsin53^@)xx9.81xx(sin53^@)/(cos53^@)#

#=>v=3.95ms^-1#, rounded to two decimal places.

(ii) Given is height of point of suspension #A# of the bob above ground #=3m#

Height of the bob from the ground
#h=3-Lcos53^@#

#=>h=3-1.5xxcos53^@=3-0.90=2.1m#

After breaking of string the bob has only one force acting on it. It falls freely under gravity. Noting that initial downward velocity #=0#, using the kinematic expression to calculate time taken by bob to land on ground

#h=ut+1/2 g t^2#

Inserting various values we get

#2.1=0xxxxt+1/2xxgxxt^2#
#=>t^2=2.1xx2/9.81#
#=>t=0.65s#, rounded to two decimal places

(iii) Horizontal distance of the bob on the ground from the rod is due to horizontal tangential velocity at the time of breaking of the string.

Horizontal Distance covered in #0.65s#

#vt=3.95xx0.65=2.57m#

Total distance on the ground from the rod
#D=sqrt((vt)^2+r^2)#
#D=sqrt(2.57^2+1.5^2xxsin^2 53^@)=2.84m#, rounded to two decimal places

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