Sample Problems on Rolling Motion

Problem 1: Three bodies, a ring, a solid cylinder, and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

Solution:

According to the conservation of energy of the rolling body, i.e. there is no loss of energy due to friction. . The potential energy lost by the body in rolling down the inclined plane is P.E= mgh, therefore, be equal to kinetic energy gained. The bodies start from rest the kinetic energy gained is equal to the final kinetic energy of the bodies. 

The expression for the kinetic energy is 

[Tex]K=\frac{1}{2}mv^2\left(\frac{k^2}{R^2}+1\right) [/Tex]

 where v is the final velocity of the Centre of mass of the body.

Equate K and mgh,

[Tex]mgh=\frac{1}{2}mv^2\left(\frac{k^2}{R^2}+1\right)\\ or\text{ }v^2=\frac{2gh}{\left(\frac{k^2}{R^2}+1\right)} [/Tex]

Here, velocity is independent of the mass of the rolling body,

For a ring, k²= R², therefore the expression can be written as

[Tex]v_{ring}^2=\frac{2gh}{\left(1+1\right)}\\ v_{ring}^2=\frac{2gh}{2}\\v_{ring}^2=gh\\ v_{ring}=\sqrt{gh} [/Tex]

For a solid cylinder k²= R²/2 , therefore,

[Tex]v_{disc}^2=\frac{2gh}{\left(\frac{1}{2}+1\right)}\\ v_{disc}^2=\frac{2gh}{\left(\frac{3}{2}\right)}\\ v_{disc}^2=\frac{4gh}{3}\\ v_{disc}=\sqrt{\frac{4gh}{3}} [/Tex]

Now for a solid sphere k²= 2R²/5 , therefore,

[Tex]v_{sphere}^2=\frac{2gh}{\left(\frac{2}{5}+1\right)}\\ v_{sphere}^2=\frac{2gh}{\left(\frac{7}{5}\right)}\\ v_{sphere}^2=\frac{10gh}{7}\\ v_{sphere}=\sqrt{\frac{10gh}{7}} [/Tex]

The results show that the sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

Problem 2: Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 MR²/5, where M is the mass of the sphere and R is the radius of the sphere. 

(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.

Solution:

 (a) 

Moment of inertia of sphere about any diameter = 2/5 MR²

Applying theorem of parallel axes

(I)Moment of inertia of sphere about a tangent to the sphere = 2/5 MR² +M(R)² 

I = 7/5 MR²

(b) 

Given,

Expression for the moment of inertia of the disc about any of its diameters can be written as 

L= 1/4 MR²

(i) Using theorem of perpendicular axes, the moment of inertia of the disc about an axis passing through its centre and normal to the disc can be written as 

I= 2 x 1/4 MR² 

I = 1/2 MR²

(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies can be written as 

I = 1/2 MR²+ MR² 

I = 3/2 MR²

Problem 3: Equal magnitude of torque is applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Solution:

 Let the radius and mass of the hollow cylinder, and the solid sphere be R and M respectively.

Moments of inertia of the hollow cylinder about the respective axes is I₁ = MR² 

Moments of inertia of the solid sphere about the respective axes I₂ = (2/5) MR²

Let the magnitude of the torque applied to the cylinder and the sphere is τ , which produce angular accelerations α₁ and α₂ respectively in the cylinder and the sphere

Magnitude of the torque can be expressed as

τ = I₁ α₁ = I₂ α₂

MR² α₁ = (2/5) MR² α₂

α₂ = 5/2 α₁

Thus, the sphere has larger angular acceleration. 

Therefore, the sphere will acquire larger angular speed after a given time.

Problem 5: A ring of mass m and radius R has three particles attached to the ring as shown in the figure. The centre of the ring has a speed v₀. If the kinetic energy of the system is xmv₀ ². Find x (Slipping is absent)

Solution:

The expression for the rolling motion when slip is absent is,

 [Tex]v_0=R\omega_0 [/Tex]

Now the speed of the 2m particle placed at left side is 

[Tex]\sqrt{v_0^2+(R\omega_0)^2}\\ \sqrt{v_0^2+v_0^2}\\ \sqrt{2}v_0 [/Tex]

The speed of the m particle placed at right side is

[Tex]\sqrt{v_0^2+(R\omega_0)^2}\\ \sqrt{v_0^2+v_0^2}\\ \sqrt{2}v_0 [/Tex]

Speed of the m particle placed at top is

[Tex]v_0+R\omega_0\\ v_0+v_0\\2v_0 [/Tex]

Kinetic energy placed at left , right and top is 

[Tex]K.E =\frac{1}{2}\times2m(\sqrt{2}v_0)^2+\frac{1}{2}\times{m}(\sqrt{2}v_0)^2+\frac{1}{2}\times{m}(2v_0)^2\\ K.E =2mv_0^2+mv_0^2+2mv_0^2\\ K.E=5mv_0^2 [/Tex]

Kinetic energy of the ring is 

K.E = Rotational Kinetic Energy + Translational Kinetic Energy

[Tex]K.E =\frac{1}{2}I\omega^2+\frac{1}{2}mv_0^2 [/Tex]

Substitute,

• I = mR²
• ω = v₀/R

[Tex]K.E =\frac{1}{2}mR^2\left(\frac{v_0}{R}\right)^2+\frac{1}{2}mv_0^2\\ K.E =mv_0^2 [/Tex]

Total Kinetic Energy is,

[Tex]K.E =5mv_0^2+mv_0^2\\ K.E =6mv_0^2 [/Tex]

Therefore, the value of x is 6.

Other Physics links:

• Data Communication
• Potential Energy
• Wave Definition Physics
• What Is Analog And Digital Signal
• Kinetic Energy Definition Physics
• Kepler’s Laws of Planetary Motion
• Application Of Electrolysis
• Schottky Diode Working
• Torque Examples

Rolling motion is one of the most relevant movements we can see in our everyday life. There is a rolling motion in all wheels used in transportation like cars, buses, trains, aero planes, bikes, and buffalo carts. We can also see the rolling motion in gears and ball bearings. In this article, we will learn about the basic fundaments of Rolling Motion like definition, examples, and its applications in the real world. So, let’s start learning the fundamental topic of Rolling Motion.