Question

A small object tied at the end of a string is to be whirled in a vertical circle of radius r. If its speed at the lowest point is 3- gr, then its speed at the highest point is​

Answer 1

Explanation:

A small object tied at the end of a string is to be whirled in a vertical circle of radius r. If its speed at the lowest point is 3- gr, then its speed at the highest point isA small object tied at the end of a string is to be whirled in a vertical circle of radius r. If its speed at the lowest point is 3- gr, then its speed at the highest point is

Answer 2

Given: An object is whirled in a vertical circle of radius r

            Speed of object at lowest point = √3gr

To Find : Speed of object at highest point of circle

Solution:

• For solving this question we will apply Conservation of Mechanical Energy which states that:

In a given isolated system, the sum of all conservative forces remains the same.

• At lowest point, let the object have Kinetic Energy but 0 Potential Energy, i.e, let it be our reference point. Let KE at lowest point be $KE_{1}$.
• At highest point object will have both KE and PE. Let it be $KE_{2}$ and $PE_{1}$.
• According to conservation of mechanical energy:

                                 $KE_{1}$ = $KE_{2}$ + $PE_{1}$                   (1)

• For calculating KE we will apply the formula KE = $\frac{1}{2}$ mv² and for calculating PE we will use PE = mgh. Here,

1. m is mass of body
2. v is velocity of body
3. g is acceleration due to gravity and
4. h is height above reference point.

• Substituting given values in (1)

                      $\frac{1}{2}$ m (√3gr)² =  mg(2r) + $\frac{1}{2}$ m$v^{2}$   (h= diameter of circle = 2r)

                             ⇒ $\frac{3mgr}{2}$ = 2mgr + $\frac{1}{2}$ m $v^{2}$

                            ⇒ $\frac{1}{2}$ m $v^{2}$ = $\frac{1}{2}$ mgr

                            ⇒ $v^{2}$ = gr

                           ⇒ v =√gr m/s

Hence, velocity of body at highest point will be $\sqrt{gr}$ m/s.