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“If a particle of mass m is projected from the lowest point of smooth vertical circle of radius 'A' with velocity √(5ag) the force at its lowest point will be”
- a)6 mg
- b)5 mg
- c)4 mg
- d)3 mg
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
“If a particle of mass m is projected from the lowest point of s...
At lowest point centrifugal force is equal to mv2/a in downward direction
=> total downward force = mg + mv2/a
= mg + m 5ag/a
= mg + 5 mg
= 6mg
=> total downward force = mg + mv2/a
= mg + m 5ag/a
= mg + 5 mg
= 6mg
Most Upvoted Answer
“If a particle of mass m is projected from the lowest point of s...
At lowest point centrifugal force is equal to mv^2/a in downward direction
=> total downward force = mg + mv^2/a
= mg + m 5ag/a
= mg + 5 mg
= 6mg
PS: v must be √5ag
=> total downward force = mg + mv^2/a
= mg + m 5ag/a
= mg + 5 mg
= 6mg
PS: v must be √5ag
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Community Answer
“If a particle of mass m is projected from the lowest point of s...
- **Given data**
Given, mass of particle, \(m\), velocity of particle, \(\sqrt{5ag}\), and radius of circle, \(A\).
- **Centripetal Force**
At the lowest point of the vertical circle, the force acting on the particle is the centripetal force required to keep the particle moving in a circular path. This force is provided by the component of gravity acting along the radius of the circle.
- **Calculation**
The velocity of the particle at the lowest point is equal to the circular velocity. Thus, the centripetal force required is given by:
\[F_c = \frac{mv^2}{r}\]
\[F_c = \frac{m(\sqrt{5ag})^2}{A}\]
\[F_c = \frac{5ma}{A}\]
- **Weight of the Particle**
At the lowest point, the weight of the particle acts downwards. The weight is given by:
\[W = mg\]
- **Comparison**
Comparing the centripetal force with the weight of the particle:
\[F_c = \frac{5ma}{A} = 5mg\]
Since the centripetal force is equal to the weight of the particle at the lowest point, the correct answer is option **A) 6mg**.
Given, mass of particle, \(m\), velocity of particle, \(\sqrt{5ag}\), and radius of circle, \(A\).
- **Centripetal Force**
At the lowest point of the vertical circle, the force acting on the particle is the centripetal force required to keep the particle moving in a circular path. This force is provided by the component of gravity acting along the radius of the circle.
- **Calculation**
The velocity of the particle at the lowest point is equal to the circular velocity. Thus, the centripetal force required is given by:
\[F_c = \frac{mv^2}{r}\]
\[F_c = \frac{m(\sqrt{5ag})^2}{A}\]
\[F_c = \frac{5ma}{A}\]
- **Weight of the Particle**
At the lowest point, the weight of the particle acts downwards. The weight is given by:
\[W = mg\]
- **Comparison**
Comparing the centripetal force with the weight of the particle:
\[F_c = \frac{5ma}{A} = 5mg\]
Since the centripetal force is equal to the weight of the particle at the lowest point, the correct answer is option **A) 6mg**.
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