Home
Class 12
PHYSICS
A weightless thread can bear tension ...

A weightless thread can bear tension upto 3.7 kg wt .A stone of mass 500 gm is tied to it and rotated in a circular path of radius 4 m, in a vertical circle If ` g=10 m//s^(2)` then the maximum angular velocity of the stone will be

A

2 rad /s

B

5 rad /s

C

4rad /s

D

3 rad /s

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow these main points: ### Step 1: Convert the tension from kg to Newtons The maximum tension that the thread can bear is given as 3.7 kg wt. To convert this into Newtons, we use the formula: \[ T = m \cdot g \] where \( m = 3.7 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \). Calculating the tension: \[ T = 3.7 \, \text{kg} \times 10 \, \text{m/s}^2 = 37 \, \text{N} \] ### Step 2: Convert the mass of the stone to kg The mass of the stone is given as 500 g. We convert this to kilograms: \[ m = 500 \, \text{g} = 0.5 \, \text{kg} \] ### Step 3: Identify the forces acting on the stone at the bottom of the vertical circle At the bottom of the vertical circle, the forces acting on the stone are: 1. The weight of the stone acting downwards: \( W = m \cdot g = 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \) 2. The tension in the thread acting upwards. ### Step 4: Apply the centripetal force equation At the bottom of the circle, the net force providing the centripetal acceleration is given by: \[ T - W = \frac{m v^2}{r} \] Where: - \( T = 37 \, \text{N} \) - \( W = 5 \, \text{N} \) - \( m = 0.5 \, \text{kg} \) - \( r = 4 \, \text{m} \) Substituting the values: \[ 37 \, \text{N} - 5 \, \text{N} = \frac{0.5 \, \text{kg} \cdot v^2}{4 \, \text{m}} \] ### Step 5: Solve for \( v^2 \) This simplifies to: \[ 32 \, \text{N} = \frac{0.5 \, v^2}{4} \] Multiplying both sides by 4: \[ 128 = 0.5 v^2 \] Now, divide by 0.5: \[ v^2 = 256 \] Taking the square root: \[ v = \sqrt{256} = 16 \, \text{m/s} \] ### Step 6: Calculate the angular velocity \( \omega \) The relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by: \[ v = r \cdot \omega \] Rearranging gives: \[ \omega = \frac{v}{r} \] Substituting the values: \[ \omega = \frac{16 \, \text{m/s}}{4 \, \text{m}} = 4 \, \text{rad/s} \] ### Conclusion The maximum angular velocity of the stone is: \[ \omega = 4 \, \text{rad/s} \]

To solve the problem step by step, we will follow these main points: ### Step 1: Convert the tension from kg to Newtons The maximum tension that the thread can bear is given as 3.7 kg wt. To convert this into Newtons, we use the formula: \[ T = m \cdot g \] where \( m = 3.7 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \). Calculating the tension: ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CIRCULAR MOTION

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP-19|1 Videos
  • CIRCULAR MOTION

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP-20|1 Videos
  • CIRCULAR MOTION

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP-17|1 Videos
  • ATOMS, MOLECULES AND NUCLEI

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|30 Videos
  • COMMUNICATION SYSTEMS

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP -20|10 Videos

Similar Questions

Explore conceptually related problems

A weightless thread can bear tension upto 3.7 kg wt A stone of mass 500g is tied to it and revolves in a verticle circle of radius 4m What will be the maximum angular velocity of the stone if g = 10 m//s^(2) .

The tensile strength of a weightless thread is 4 kg wt . Consider a stone of mass 200 g is tied to it and evolves in a vertical plane of radius 5 m . Calculate the maximum angular velocity of the stone .

Knowledge Check

  • A weightless thread can bear tension upto 37 N . A stone of mass 500 g is tied to it and revolved in a circular path of radius 4 m in a vertical plane . If g =10 ms^(-2) , then the maximum angular velocity of the stone will be

    A
    `2rad s^(-1)`
    B
    `4rad s^(-1)`
    C
    `8rad s^(-1)`
    D
    `16rad s^(-1)`
  • A weightless thread can bear tension upto upto 3.7 Kg weight A stone of mass 500 gram is tied at its one end and revolved in a vertical circular path of radius 4 m .If g= 10 m//s^(2) then the maximum angular velocity of the stone in ( radians / sec ) will be

    A
    3
    B
    4
    C
    5
    D
    6
  • A weightless thread can bear tension upto 3.2 kg weight. A stone of mass 500 g is tied to it and removed in a circular path of radius 4 m is vertical plane. If g=10^(2)m//s^(2) , the maximum agular velocity of the stone is

    A
    2 rad/s
    B
    `sqrt(21)` rad/s
    C
    16 rad/s
    D
    4 rad/s
  • Similar Questions

    Explore conceptually related problems

    A weightless thread can bear tension up to 3.7 kg weight. A stone of mass 500 g is tied at its one end revolved in a vertical circular path of radius 4m. If g=10m//s^(2) , then the maximum angular velocity of the stone is (rad/s) is

    A weightless thread can with stand tension upto 30N.A stone of mass 0.5 kg is tied to it and is revolved in a circular path of radius 2 m in a vertical plane. If g=10 ms^(-2) , then maximum angular velocity of stone is

    A weightless thread can support tension up to 30N.A particle of mass 0.5kg is tied to it and is revolved in a circle of radius 2m in a verticle plane. If g=10m//s^(2) , then the maximum angular velocity of the stone will be

    If a stone of mass m is rotated in a vertical circular path of radius 1m, the critical velocity is

    If a particle of mass 1 gm is moving along a circular path of radius 1 m with a velocity of 1 m/s, then the its angular momentum is