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Q7.19:
A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Solution :

Step 1: Identifying Given Variables and Unit Conversion

To ensure consistency across the International System of Units (SI), we convert all values to meters ($m$), kilograms ($kg$), and seconds ($s$).

  • Radius of the hoop ($R$) = $2\text{ m}$
  • Mass of the hoop ($M$) = $100\text{ kg}$
  • Linear velocity of the centre of mass ($v_{cm}$) = $20\text{ cm/s} = 0.2\text{ m/s}$

Step 2: Theoretical Basis for Work Done

Per the Work-Energy Theorem, the work required to stop a moving object is equal to the negative of its total kinetic energy ($K_{total}$). Since the hoop is undergoing rolling motion (rotation plus translation), the total kinetic energy is the sum of translational kinetic energy ($K_t$) and rotational kinetic energy ($K_r$).

$|W| = K_{total} = K_t + K_r$

$W = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2$

Step 3: Calculating Moment of Inertia and Angular Velocity

For a thin hoop (or ring) rotating about its central axis, the moment of inertia ($I$) is defined as:

$I = MR^2$ [Per the definition of the moment of inertia for a thin hoop]

For pure rolling motion without slipping, the relationship between linear velocity and angular velocity ($\omega$) is:

$\omega = \frac{v_{cm}}{R}$

Step 4: Algebraic Derivation of Total Energy

Substitute the expressions for $I$ and $\omega$ into the energy equation:

$K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}(MR^2)\left(\frac{v_{cm}}{R}\right)^2$

$K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}(MR^2)\left(\frac{v_{cm}^2}{R^2}\right)$

$K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}Mv_{cm}^2 = Mv_{cm}^2$

Step 5: Final Substitution and Calculation

Using the values derived in Step 1:

$W = 100\text{ kg} \times (0.2\text{ m/s})^2$

$W = 100 \times 0.04$

$W = 4\text{ J}$

R=2m v=0.2m/s CM

Final Solution: The work that must be done to stop the hoop is 4 Joules.


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