Rotational Energy

Rotational Energy with Otters

Kinetic Energy

  • Translational Kinetic Energy (K_T) is given by \frac{1}{2} mv^2 . When a tennis ball is tossed, the total mechanical energy (ME) is the sum of potential energy and translational kinetic energy: ME = mgh + \frac{1}{2} mv^2.

  • If the ball is spinning, it possesses rotational kinetic energy as well; each tiny mass within the ball rotates around the axis of rotation and thus has kinetic energy.

Rotational Kinetic Energy

  • Rotational Kinetic Energy (KR) is analogous to translational kinetic energy: KR = \frac{1}{2} I\omega^2, where I is the moment of inertia and \omega is the angular velocity.

  • The total kinetic energy of a moving and spinning object is the sum of its translational and rotational kinetic energies: KT + KR = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2.

  • The total mechanical energy of such an object is ME = mgh + \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2.

Kinetic Energy of a Boomerang

  • Boomerangs require spin to function, implying that work is done to impart rotational kinetic energy to them. Both translational and rotational kinetic energy components exist.

  • Example: A 1 kg boomerang is thrown with an initial velocity of 25 m/s at an angle of 40° and an angular velocity of 100 rad/s. The moment of inertia is I = 0.05 kg \cdot m^2.

  • Translational kinetic energy: K_T = \frac{1}{2} mv^2 = \frac{1}{2} (1)(25)^2 = 312.5 J

  • Rotational kinetic energy: K_R = \frac{1}{2} I\omega^2 = \frac{1}{2} (0.05)(100)^2 = 250 J

  • Total mechanical energy: ME = KT + KR = 312.5 + 250 = 562.5 J

Conservation of Mechanical Energy in Boomerang Flight

  • As the boomerang rises, mechanical energy is conserved, with kinetic energy being converted into gravitational potential energy (U_g).

  • Gravity acts on the center of mass, so it doesn't apply torque; absent torque, rotational motion remains uniform, similar to uniform horizontal motion in free fall (no horizontal force).

  • At the peak of its trajectory, ME = mgh + KR + \frac{1}{2} mvx^2.

  • The loss in translational kinetic energy equals the gain in gravitational potential energy. Thus, \frac{1}{2} mvy^2 = mgh , leading to h = \frac{vy^2}{2g} . Given v_y = 25 \cdot sin(40) = 16.12 m/s, then h = \frac{16.12^2}{2 \cdot 9.8} \approx 13 m.

Rolling Without Slipping

  • The angular velocity (\omega) and linear velocity (v) are not always related. Slipping or sliding occurs when there's no direct correlation between \omega and v.

  • Rolling without slipping is a unique case where the point of contact between the rolling object and the surface is instantaneously at rest (static).

Relationship Between Linear and Angular Velocity

  • For rolling without slipping, the relationship between linear and angular velocity is v = r\omega, where r is the radius of the rolling object. This condition is valid only when the object is rolling without any slippage.

Ramps and Friction

  • For an object to roll down a ramp, a torque must be applied to initiate rotation. This torque is generated by a tangential force acting on the ball.

  • Friction between the ball and the ramp can provide this torque.

  • The force of gravity component along the ramp (mg\sin(\theta)) pulls the ball down, while friction opposes this motion, acting upwards along the ramp.

  • The torque due to friction is given by \tau = r \cdot f , where f is the force of friction and r the radius, creating a clockwise torque.

  • On a frictionless ramp, a ball will slide without rotating because there's no frictional force to initiate rotation.

Rolling Up a Ramp

  • As a ball rolls up a ramp, its kinetic energy is converted into gravitational potential energy.

  • On a frictionless ramp, without torque, there's no change in angular velocity, and the ball keeps spinning at a constant rate.

  • In reality, due to friction, a torque exists which causes the ball to lose both translational and rotational velocity as it ascends the ramp. Eventually, the ball comes to a complete stop, both translationally and rotationally.

Rolling Down a No-Slip Surface and Half-Pipe

  • As a ball rolls down a ramp with friction, it gains both translational and rotational kinetic energy.

  • At the bottom, the gravitational potential energy is converted into a combination of translational and rotational kinetic energy.

  • In a half-pipe scenario, consider a ball released from rest on a no-slip surface, transitioning to a frictionless surface after the lowest point.

Half-Pipe Dynamics

  • Only translational kinetic energy is converted back into potential energy on the frictionless side. Since the ball retains some rotational kinetic energy, it won't reach the same height from which it was released.

  • On the frictionless side, the ball maintains constant rotational kinetic energy, which doesn't contribute to increasing gravitational potential energy.

Rolling Hoop

  • Consider a 2 kg hoop rolling without slipping across a table at 8 m/s. What is its total kinetic energy?

  • The total kinetic energy comprises both translational and rotational kinetic energy.

  • KE = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2, where I = mr^2 for a hoop and \omega = \frac{v}{r} (rolling without slipping).

  • Even without knowing the radius, the problem can be solved because the rotational kinetic energy will be related to the translational kinetic energy.

Kinetic Energy Calculation for Rolling Hoop

  • Rewriting the rotational kinetic energy term: \frac{1}{2} I\omega^2 = \frac{1}{2} (mr^2)(\frac{v}{r})^2 = \frac{1}{2} mv^2.

  • The radius squared cancels out. Hence, both translational and rotational kinetic energies are equal in magnitude for a hoop.

  • Total KE = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 = mv^2 = (2)(8)^2 = 128 J.

Other Rolling Objects (Disk and Ball)

  • Unlike a hoop, for a disk with the same mass and velocity, I = \frac{1}{2} mr^2, and \omega = \frac{v}{r} (rolling without slipping).

  • K_R = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} mr^2)(\frac{v}{r})^2 = \frac{1}{4} mv^2.

  • Total KE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 = \frac{3}{4} (2)(8)^2 = 96 J.

  • For a solid ball, KR = \frac{2}{5} KT. If KT = \frac{1}{2} mv^2 = \frac{1}{2} (2)(8)^2 = 64 J, then KR = \frac{2}{5} (64) = 25.6 J.

Ramps and Energy Conservation

  • Always consider Conservation of Energy, i.e., U_g \rightarrow KE, and the effect of friction on both translational and rotational kinetic energies.

  • An object that slides (no rotation) reaches the bottom faster with a greater final speed because all potential energy converts into translational kinetic energy.

  • Energy diverted into rotational kinetic energy reduces the final translational velocity. Ug = KT + K_R = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2.

Race to the Bottom

  • A solid sphere and a hollow sphere (same mass and radius) are released simultaneously on a ramp. Which reaches the bottom first?

  • Both start with the same potential energy, but the distribution between translational and rotational kinetic energy differs.

  • The sphere that loses the least energy to rotational kinetic energy will have more translational kinetic energy and arrive first.

Moment of Inertia and Rotational Kinetic Energy

  • The difference in rotational kinetic energy stems from the moment of inertia coefficients.

  • For a solid sphere, KR = \frac{1}{2} (\frac{2}{5} mv^2). For a hollow sphere, KR = \frac{1}{2} (\frac{2}{3} mv^2).

  • A smaller coefficient means less rotational kinetic energy, more translational kinetic energy, and a higher final velocity.

  • The solid sphere, having a smaller coefficient, wins the race.

The Ultimate Race

  • Comparing a block (sliding), hoop, disk, hollow ball, and solid ball racing down a ramp. The block wins because it has only translational kinetic energy.

  • A sliding block converts all Ug into KT, no K_R. and therefore has both the highest velocity and shortest time.

Order of Finish

  • The order is determined by the coefficients: solid sphere (2/5), disk (1/2), hollow sphere (2/3), and hoop (1).

Same Shape, Different Size

  • If a small, light cylinder and a large, heavy cylinder are released simultaneously, who wins?

  • The key is to determine what variables affect speed. Do m or v matter?

  • mgh = \frac{1}{2} mv^2 + \frac{1}{2} I\omega^2. We can substitute \frac{1}{2} I\omega^2 \rightarrow \frac{1}{2} (\frac{1}{2})mr^2(\frac{v}{r})^2 \rightarrow \frac{1}{4} mv^2.

  • mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 \rightarrow gh = \frac{3}{4} v^2. Both r and m cancel out, so only shape matters, and the cylinders tie.

How Far Up the Ramp?

  • If a hollow ball rolls at 6 m/s onto a 30° ramp, how far will it travel up the ramp (without slipping)?

  • KT will transform into Ug , but what about KR? Since there is no slipping, the translational and rotational motion are related, and both will be converted into Ug .

  • At the highest point, the object will have zero KT and zero KR, with all energy as mgh. So how much total KE does it start with?

Distance Calculation

  • \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{2}{3} mr^2)(\frac{v}{r})^2 = \frac{1}{3} mv^2 .

  • mgh = \frac{1}{2} mv^2 + \frac{1}{3} mv^2 \rightarrow gh = \frac{5}{6} v^2 \rightarrow h = \frac{5}{6} \frac{v^2}{g} = \frac{5(6)^2}{6(10)} = 3 m.

  • The height is 3m, but the question wants distance along the 30° ramp. So, it goes 6m up the ramp to reach a height of 3m, realizing that sin 30° = 1/2.