Question

The bob of a simple pendulum of length 1m has mass 100g and speed 1.4m/s at the lowest point in its path. Find the tension in the string at its instant

Answer 1

Answer:

Length of simple pendulum= 1m

mass = 100gm = 0.1 kg

it is displaced through an angle of 60 from the vertical

Height of pendulum at starting position= l(1-cos60) = 1(1 - 0.5) = 0.5m

Potential energy = mgh = 0.1×10×0.5 = 0.5J

When it is released and it reaches mean position, its potential energy at starting point is converted into kinetic energy.

So KE of bob at mean position = 0.5 J

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Answer 2

Answer:

Given conditions ⇒

Given conditions ⇒Mass =100 g

Given conditions ⇒Mass =100 g=0.10 kg

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/s

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 m

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force∴ Tension = mg + mv²/r

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force∴ Tension = mg + mv²/r∴ T = 0.10 × 10 + 0.10 × (1.4)²/1

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force∴ Tension = mg + mv²/r∴ T = 0.10 × 10 + 0.10 × (1.4)²/1⇒ T = 1 + 0.196 N

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force∴ Tension = mg + mv²/r∴ T = 0.10 × 10 + 0.10 × (1.4)²/1⇒ T = 1 + 0.196 N∴ T = 1.196 N

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force∴ Tension = mg + mv²/r∴ T = 0.10 × 10 + 0.10 × (1.4)²/1⇒ T = 1 + 0.196 N∴ T = 1.196 NHence, the tension in the string at the Instant = 1.196 N.

Given conditions ⇒Mass =100 g=0.10 kgVelocity (v)=1.4 m/sRadius(r)=1 mThe Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.Therefore,Tension = Force of gravity + Centripetal Force∴ Tension = mg + mv²/r∴ T = 0.10 × 10 + 0.10 × (1.4)²/1⇒ T = 1 + 0.196 N∴ T = 1.196 NHence, the tension in the string at the Instant = 1.196 N.Hope it helps.