Question

A ball of mass 0.3 kg is tied to one end of a string 0.8 m long and revolved in a vertical circle.At what speed of the ball will the tension in the string be zero at the highest point of the circle?What would be the tension at the lowest point in this situation ?
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Answer 1

Answer:

2.8 m/s     ; 17.64 N

Explanation:

At the highest point,  

centripetal force = tension(string) + gravity

⇒ mv²/r = T + mg

 Since, tension at the stage is 0,

⇒ mv²/r = mg

⇒ v² = rg

⇒ v = √(0.8)(9.8) = 2.8 m/s

When the velocity of the mass at the highest point is 2.8, total energy was kinetic energy + potential = 1/2 mv² + mgh  

At a distance 2r(=2*0.8) from ground, total energy = 1/2 mv² + mg(1.6)

 ∴ energy = 1/2 mv² + 1.8mg

                  = 5.88 joule

At the lowest point, potential energy is 0, and kinetic energy is 1/2 mu², where u is the velocity at the point.  

⇒ 1/2 mu² = 5.88  ⇒ mu² = 11.76

At the lowest point,

tension = mg + mu²/r

             = (0.3)(9.8) + 11.76/0.8

             = 17.64 N

Answer 2

For tension to be zero at highest point

mv^2 / r = mg

v = sq rt (rg)

v = sq rt(0.8 * 9.8)

v = sq rt (7.84)

v = 2.8 m/s

Tension at lowest point

T = mv^2/r + mg

= m * (5gr) / r + mg

= 6mg

= 6 * 0.3 * 9.8

= 17.64 N

Tension at lowest point is 17.64 N

abhi569

abhi569

Answer:

2.4 m/s ; 17.64 N

Explanation:

At the highest point,

centripetal force = tension(string) + gravity

⇒ mv²/r = T + mg

Since, tension at the stage is 0,

⇒ mv²/r = mg

⇒ v² = rg

⇒ v = √(0.8)( 9.8) = 2.8 m/s

When the velocity of the mass at the highest point is 2.8, total energy was kinetic energy + potential = 1/2 mv² + mgh

At a distance 2r(=2*0.8) from ground, total energy = 1/2 mv² + mg(1.6)

∴ energy = 1/2 mv² + 1.8mg

= 5.88 joule

At the lowest point, potential energy is 0, and kinetic energy is 1/2 mu², where u is the velocity at the point.

⇒ 1/2 mu² = 5.88 ⇒ mu² = 11.76

$At \: the \: lowest \: point,$

tension = mg + mu²/r

= (0.3)( 9.8) + 11.76/0.8

$\: = 17.64 N$