Rotational Dynamics

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8. A uniform horizontal lever of length 2.4 m pivoted at its mid-point is in equilibrium. The following forces are acting on the lever: a 5 N downward force (force with downward vertical component) that makes an angle of 30 deg with the horizontal-left acting at the right end of the lever, an unknown vertically downward force acting at the left end of the lever, and a 10 N vertically downward force acting at a point on the lever 0.7 m away from the right end.

Calculate the magnitude of the unknown force.

A. 4.667 N B. 6.667 N C. 9.333 N D. 7.333 N E. 8 N

9. Three particles of masses 4 kg, 7 kg and 8 kg are located at the points (0, 1) m, (16, -4) m, and (2, -3) m respectively. Determine the location of the center of gravity of the three particles.

A. (7.411, -2.274) m B. (7.411, -2.526) m C. (6.737, -2.274) m D. (8.084, -2.021) m E. (6.737, -2.526) m

10. The center of gravity of two particles of masses 18 kg and 11 kg is located at the point (4, -1) m.

The 18 kg particle is located at the point (4, 3) m. Find the location of the 11 kg particle.

A. (3.2, -7.545) m B. (2.8, -4.527) m C. (4, -7.545) m D. (4, -8.3) m E. (3.2, -8.3) m

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The product mr⊥ 2 is called the moment of inertial of the object about the given axis and denoted by I.

I = mr⊥ 2

The unit of measurement for moment of inertia is kg m ². The relationship between torque and angular acceleration is that torque is proportional to angular acceleration with the constant of proportionality being the moment of inertia of the object about the given axis. Moment of inertia of an object is a constant for a given axis of rotation but different for different axes of rotation.

τ = Iα

Example: An object of mass 5 kg is revolving in a circular path of radius 2 m with an angular acceleration of 10 rad ⁄ s ².

a) Calculate its moment of inertia about an axis passing through the center of the circle perpendicularly.

Solution: m = 5 kg; r⊥ = 2 m; I = ?

I = mr⊥ 2 = 5 * 2 kg m ² = 20 kg m ²

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b) Calculate the torque acting on the object about an axis that passes through the center of the circle perpendicularly.

Solution: α = 10 rad ⁄ s ²; τ = ?

τ = Iα = 20 * 10 N m = 200 N m

For a system involving more than one particle, the moment of inertia of the system is equal to the sum of the moment of inertias of the individual particles.

I = ∑i mi ri⊥ 2 = m1 r1⊥ 2 + m2 r2⊥ 2 + …

Example: A system consists of three particles. The first has a mass of 6 kg and is located at (2, 3) m.

The second has a mass of 4 kg and is located at (-2, 4) m. The third has a mass of 5 kg and is located at (0, -3) m.

a) Calculate the moment of inertia of the system about the x-axis.

Solution: If the x-axis is the axis of rotation, then the perpendicular distance between the axis and a particle is equal to the absolute value of the y-coordinate of the particle.

m1 = 6 kg; r1⊥ = 3 m; m2 = 4 kg; r2⊥ = 4 m; m3 = 5 kg; r3⊥ = 3 m; Ix = ? Ix = m1 r1⊥ 2 + m2 r2⊥ 2 + m3 r3⊥ 2 = (6 * 3 ² + 4 * 4 ² + 5 * 3 ²) kg m ² = 163 kg m ² b) Calculate the moment of inertia of the system about the y-axis.

Solution: If the y-axis is the axis of rotation, then the perpendicular distance between the axis and a particle is equal to the absolute value of the x-coordinate of the particle.

m1 = 6 kg; r1⊥ = 2 m; m2 = 4 kg; r2⊥ = 2 m; m3 = 5 kg; r3⊥ = 0 m; Ix = ? Iy = m1 r1⊥ 2 + m2 r2⊥ 2 + m3 r3 ⊥2 = (6 * 2 ² + 4 * 2 ² + 5 * 0 ²) kg m ² = 40 kg m ²

c) If the system is rotating about the y-axis with an angular acceleration of 7 rad ⁄ s ², calculate the torque acting on these system of particles.

Solution: Iy = 40 kg m ²; α = 7 rad ⁄ s ²; τ = ?

τ = Iy α = 40 * 7 N m = 280 N m

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Obtaining moment of inertia of solid objects requires the use of calculus. But formulas for the moment of inertia of common shapes such as cylinder and sphere may be found in physics text books. For example the moment of inertia of a spherical object of mass M and radius R about an axis through the center of the sphere is given by Isphere = 2MR ² ⁄ 5.

8.4.2 Rotational Kinetic Energy

Let’s consider an object of mass m revolving in a circular path of radius r⊥ with a speed of v. This motion can be looked at as linear motion in the circular trajectory or rotational motion about the center. Its linear kinetic energy which should be equal to the rotational kinetic energy is given by KE = mv ² ⁄ 2. But the linear speed v can be expressed in terms of the angular speed ω as v = r⊥ ω; and the kinetic energy may be written as KE = mr⊥ 2ω ² ⁄ 2. The expression mr⊥ 2 is the moment of inertia, I, of the object. Therefore the rotational kinetic energy (KErot) of the object is given by the following expression.

KErot = Iω ² ⁄ 2

Example: A solid spherical object of mass 10 kg and radius 0.2 kg is rotating about an axis that passes through its center with an angular speed of 2 rad ⁄ s. Calculate its rotational kinetic energy.

Solution: M = 10 kg; R = 0.2 m; ω = 2 rad ⁄ s; KErot = ?

Isphere = 2MR ² ⁄ 5 = 2 * 10 * 0.2 ² ⁄ 5 kg m ² = 1.6 kg m ² KErot = Isphere ω ² ⁄ 2 = 1.6 * 2 ² ⁄ 2 J = 3.2 J 8.4.3 Kinetic Energy of a Rolling Object

A rolling object has both translational and rotational kinetic energy because it rotates as it moves. The kinetic energy of a rolling object is the sum of its translational (KEtra) and rotational kinetic energy (KErot). If a rolling object is moving with a speed of v and rotating with angular speed ω, its kinetic energy (KErol) is given by

KErol = KEtra + KErot = Mv ² ⁄ 2 + Iω ² ⁄ 2

The translational speed, v, and rotational speed, ω, are related: v = Rω. R is the radius of rotation.

Example: A sphere of mass 5 kg and radius 0.1 m is rolling in a horizontal surface with a speed of 4 m ⁄ s. Calculate its kinetic energy.

Solution: M = 5 kg; R = 0.1 m; v = 4 m ⁄ s; KErol = ?

Isphere = 2MR ² ⁄ 5 = 2 * 5 * 0.1 ² kg m ² ⁄ 5 = 0.02 kg m ²

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ω = v ⁄ R = 4 ⁄ 0.1 rad ⁄ s = 40 rad ⁄ s

KErol = Mv ² ⁄ 2 + Isphere ω ² ⁄ 2 = 5 * 4 ² ⁄ 2 + 0.02 * 40 ² ⁄ 2 J = 48 J

Example: A spherical object of radius 0.2 m is rolling down a 10 m 30° inclined plane. Calculate its speed by the time it reaches the ground.

Solution: The forces acting on the object are gravity and friction. Even though friction is non-conservative, it doesn’t contribute to the work done because each particle of the sphere is only instantaneously in contact with the plane. Thus the principle of conservation of mechanical energy can be applied. Let the origin of the coordinate system be fixed at the ground.

yf = 0; vi = ωi = 0; yi = 10 * sin 30° m = 5 m; R = 0.2; vf = ?

mvi 2 ⁄ 2 + Iωi 2 ⁄ 2 + m|g|yi = mvf 2 ⁄ 2 + Iωf 2 ⁄ 2 + m|g|yf

m|g|yi = mvf 2 ⁄ 2 + (2mR ² ⁄ 5)(vf ⁄ R) ² ⁄ 2 vf = √(10|g|yi ⁄ 7) = √(10 * 9.8 * 5 ⁄ 7) m ⁄ s = 8.4 m ⁄ s

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In document 360° thinking . (Page 158-163)