A cylindrical hoop is released from rest at a height of 0.7m near the top of the inclined plane.If the hoop rolls down the plane without slipping and there is no loss of energy due to friction;find the linear speed of the hoop when it reaches the bottom of the plane. Given moment of inertia of cylindrical hoop = MR ².

To find the linear speed of a cylindrical hoop when it reaches the bottom of an inclined plane, we can use the principles of conservation of energy. Since the hoop rolls without slipping and there is no energy loss due to friction, the potential energy at the top will convert entirely into kinetic energy at the bottom.

Understanding Energy Conservation

Initially, the hoop has gravitational potential energy (PE) due to its height. As it rolls down, this potential energy transforms into two forms of kinetic energy: translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its spinning).

Formulas to Use

The potential energy (PE) at the height \( h \) is given by:

• PE = mgh

Where:

• m = mass of the hoop
• g = acceleration due to gravity (approximately 9.81 m/s²)
• h = height (0.7 m in this case)

The total kinetic energy (KE) at the bottom consists of translational and rotational components:

• KE = KE_translational + KE_rotational
• KE_translational = \(\frac{1}{2} mv^2\)
• KE_rotational = \(\frac{1}{2} I \omega^2\)

For a cylindrical hoop, the moment of inertia \( I \) is given by:

• I = MR²

Since the hoop rolls without slipping, the relationship between linear speed \( v \) and angular speed \( \omega \) is:

• \( v = R\omega \)

Setting Up the Equation

At the top of the incline, the total energy is purely potential:

• Initial Energy = mgh

At the bottom, the total energy is the sum of translational and rotational kinetic energies:

• Final Energy = \(\frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\)

Substituting the moment of inertia and the relationship between \( v \) and \( \omega \), we get:

• Final Energy = \(\frac{1}{2} mv^2 + \frac{1}{2} (MR^2)(\frac{v}{R})^2\)

This simplifies to:

• Final Energy = \(\frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2\)

Equating Energies

Now we can set the initial potential energy equal to the final kinetic energy:

• mgh = mv^2

Notice that the mass \( m \) cancels out from both sides:

• gh = v^2

Now, we can solve for \( v \):

• v = \sqrt{gh}

Calculating the Final Speed

Substituting the values for \( g \) and \( h \):

• g = 9.81 m/s²
• h = 0.7 m

Now plug these into the equation:

• v = \sqrt{9.81 \times 0.7}

Calculating this gives:

• v = \sqrt{6.867} ≈ 2.62 m/s

Final Result

The linear speed of the hoop when it reaches the bottom of the inclined plane is approximately 2.62 m/s. This demonstrates how energy conservation principles can be applied to solve problems involving rolling motion.