Rotational Kinematics and Energy

Rotational Motion Up until now we have been looking at the kinematics and dynamics of translational motion – that is, motion without rotation. Now we will widen our view of the natural world to include objects that both rotate and translate. We will develop descriptions (equations) that describe rotational motion Now we can look at motion of bicycle wheels, roundabouts and divers.

Rotational variables - Angular position, displacement, velocity, acceleration II. Rotation with constant angular acceleration III. Relation between linear and angular variables - Position, speed, acceleration IV. Kinetic energy of rotation V. Rotational inertia VI. Torque VII. Newton’s second law for rotation VIII. Work and rotational kinetic energy

Rotational kinematics In the kinematics of rotation we encounter new kinematic quantities Angular displacement q Angular speed w Angular acceleration a Rotational Inertia I Torque t All these quantities are defined relative to an axis of rotation

Angular displacement Measured in radians or degrees There is no dimension Dq = qf - qi qi Dq Axis of rotation qf

Angular displacement and arc length Arc length depends on the distance it is measured away from the axis of rotation Axis of rotation Q sp sq P r qi qf

Angular Speed Angular speed is the rate of change of angular position We can also define the instantaneous angular speed

Average angular velocity and tangential speed Recall that speed is distance divided by time elapsed Tangential speed is arc length divided by time elapsed And because we can write

Average Angular Acceleration Rate of change of angular velocity Instantaneous angular acceleration

Angular acceleration and tangential acceleration We can find a link between tangential acceleration at and angular acceleration α So

Centripetal acceleration We have that But we also know that So we can also say

Example: Rotation A dryer rotates at 120 rpm. What distance do your clothes travel during one half hour of drying time in a 70 cm diameter dryer? What angle is swept out?

Rotational motion with constant angular acceleration We will consider cases where a is constant Definitions of rotational and translational quantities look similar The kinematic equations describing rotational motion also look similar Each of the translational kinematic equations has a rotational analogue

Rotational and Translational Kinematic Equations

Constant a motion What is the angular acceleration of a car’s wheels (radius 25 cm) when a car accelerates from 2 m/s to 5 m/s in 8 seconds?

A rotating wheel requires 3. 00 s to rotate through 37. 0 revolutions A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00-s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel?

Rotational Dynamics Easier to move door at A than at B using the same force F More torque is exerted at A than at B hinge A B

Torque Torque is the rotational analogue of Force Torque, t, is defined to be Where F is the force applied tangent to the rotation and r is the distance from the axis of rotation F r

Torque t = Fsinq r A general definition of torque is F Units of torque are Nm Sign convention used with torque Torque is positive if object tends to rotate CCW Torque is negative if object tends to rotate CW r F q t = Fsinq r

Condition for Equilibrium We know that if an object is in (translational) equilibrium then it does not accelerate. We can say that SF = 0 An object in rotational equilibrium does not change its rotational speed. In this case we can say that there is no net torque or in other words that: St = 0

Torque and angular acceleration An unbalanced torque (t) gives rise to an angular acceleration (a) We can find an expression analogous to F = ma that relates t and a We can see that Ft = mat and Ftr = matr = mr2a (since at = ra) Therefore Ft r m t = mr2a

Torque and Angular Acceleration t = mr2a Angular acceleration is directly proportional to the net torque, but the constant of proportionality has to do with both the mass of the object and the distance of the object from the axis of rotation – in this case the constant is mr2 This constant is called the moment of inertia. Its symbol is I, and its units are kgm2 I depends on the arrangement of the rotating system. It might be different when the same mass is rotating about a different axis

Newton’s Second Law for Rotation We now have that Where I is a constant related to the distribution of mass in the rotating system This is a new version of Newton’s second law that applies to rotation t = Ia

A fish takes a line and pulls it with a tension of 15 N for 20 seconds A fish takes a line and pulls it with a tension of 15 N for 20 seconds. The spool has a radius of 7.5 cm. If the moment of inertia of the reel is 10 kgm2, through how many rotations does the reel spin? (Assume there is no friction)

Angular Acceleration and I The angular acceleration reached by a rotating object depends on, M, r, (their distribution) and When objects are rolling under the influence of gravity, only the mass distribution and the radius are important •

Moments of Inertia for Rotating Objects I for a small mass m rotating about a point a distance r away is mr2 What is the moment of inertia for an object that is rotating – such as a rolling object? Disc? Sphere? Hoop? Cylinder?

Moments of Inertia for Rotating Objects The total torque on a rotating system is the sum of the torques acting on all particles of the system about the axis of rotation – and since a is the same for all particles: I  Smr2 = m1r12+ m2r22+ m3r32+… Axis of rotation

Continuous Objects To calculate the moment of inertia for continuous objects, we imagine the object to consist of a continuum of very small mass elements dm. Thus the finite sum Σmi r2i becomes the integral

Moment of Inertia of a Uniform Rod Find the moment of inertia of a uniform rod of length L and mass M about an axis perpendicular to the rod and through one end. Assume that the rod has negligible thickness. L

Example:Moment of Inertia of a Dumbbell A dumbbell consist of point masses 2kg and 1kg attached by a rigid massless rod of length 0.6m. Calculate the rotational inertia of the dumbbell (a) about the axis going through the center of the mass and (b) going through the 2kg mass.

Example:Moment of Inertia of a Dumbbell

Moment of Inertia of a Uniform Hoop dm All mass of the hoop M is at distance r = R from the axis R

Moment of Inertia of a Uniform Disc We expect that I will be smaller than MR2 since the mass is uniformly distributed from r = 0 to r = R rather than being concentrated at r = R as it is in the hoop. dr r R Each mass element is a hoop of radius r and thickness dr. Mass per unit area σ = M / A = M /πR2

Moment of Inertia of a Uniform Disc dr r R

Moments of inertia I for Different Mass Arrangements

Moments of inertia I for Different Mass Arrangements

Which one will win? A hoop, disc and sphere are all rolled down an inclined plane. Which one will win?

Which one will win? A hoop, disc and sphere are all rolled down an inclined plane. Which one will win? 1. Hoop I = MR2 2. Disc I = ½MR2 3. Sphere I = 2/5MR2 = t / I a1 = t / MR2 α2= 2(t / MR2) 3. a3 = 2.5(t / MR2)

Kinetic energy of rotation Reminder: Angular velocity, ω is the same for all particles within the rotating body. Linear velocity, v of a particle within the rigid body depends on the particle’s distance to the rotation axis (r). Moment of Inertia

Kinetic energy of a body in pure rotation Kinetic energy of a body in pure translation

VII. Work and Rotational kinetic energy Translation Rotation Work-kinetic energy Theorem Work, rotation about fixed axis Work, constant torque Power, rotation about fixed axis Proof:

Rotational Kinetic Energy We must rewrite our statements of conservation of mechanical energy to include KEr Must now allow that (in general): ½ mv2+mgh+ ½ Iw2 = constant Could also add in e.g. spring PE

Example - Rotational KE What is the linear speed of a ball with radius 1 cm when it reaches the end of a 2.0 m high 30o incline? mgh+ ½ mv2+ ½ Iw2 = constant Is there enough information? 2 m

Example - Rotational KE mgh+ ½ mv2+ ½ Iω2 = constant So we have that: The velocity of the centre of mass and the tangential speed of the sphere are the same, so we can say that: Rearranging for vf:

Example: Conservation of KEr A boy of mass 30 kg is flung off the edge of a roundabout (m = 400 kg, r = 1 m) that is travelling at 2 rpm. What is the speed of the roundabout after he falls off?

During a certain period of time, the angular position of a swinging door is described by θ= 5.00 + 10.0t + 2.00t2 where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) at t = 3.00 s.

The four particles are connected by rigid rods of negligible mass The four particles are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. If the system rotates in the xy plane about the z axis with an angular speed of 6.00 rad/s, calculate (a) the moment of inertia of the system about the z axis and (b) the rotational kinetic energy of the system.

Parallel axis theorem R Rotational inertia about a given axis = Rotational Inertia about a parallel axis that extends trough body’s Center of Mass + Mh2 h = perpendicular distance between the given axis and axis through COM. Proof:

Many machines employ cams for various purposes, such as opening and closing valves. In Figure, the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed ω about the axis of the shaft?

Find the net torque on the wheel in Figure about the axle through O if a = 10.0 cm and b = 25.0 cm.

A block of mass m1 = 2. 00 kg and a block of mass m2 = 6 A block of mass m1 = 2.00 kg and a block of mass m2 = 6.00 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle = 30.0 as in Figure. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. Determine (a) the acceleration of the two blocks, and (b) the tensions in the string on both sides of the pulley.

A uniform rod 1. 1 m long with mass 0 A uniform rod 1.1 m long with mass 0.7 kg is pivoted at one end, as shown in Fig., and released from a horizontal position. Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction.

A seesaw pivots as shown in Fig A seesaw pivots as shown in Fig. (a) What is the net torque about the pivot point? (b) Give an example for which the application of three different forces and their points of application will balance the seesaw. Two of the forces must point down and the other one up.

Four small spheres are fastened to the corners of a frame of negligible mass lying in the xy plane (Fig. 10.7). Two of the spheres have mass m = 3.1kg and are a distance a = 1.7 m from the origin and the other two have mass M = 1.4 kg and are a distance a = 1.5 m from the origin. (a) If the rotation of the system occurs about the y axis, as in Figure a, with an angular speed ω = 5.1rad/s, find the moment of inertia Iy about the y axis and the rotational kinetic energy about this axis. Suppose the system rotates in the xy plane about an axis (the z axis) through O (Fig. b). Calculate the moment of inertia about the z axis and the rotational energy about this axis.

The reel shown in Figure has radius R and moment of inertia I The reel shown in Figure has radius R and moment of inertia I. One end of the block of mass m is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and is then released from rest. (a) Find the angular speed of the reel when the spring is again unstretched. (b) Evaluate the angular speed numerically at this point if I = 1.00 kg·m2, R = 0.300 m, k = 50.0 N/m, m = 0.500 kg, d = 0.200 m, and θ= 37.0°.

A tennis ball is a hollow sphere with a thin wall A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a horizontal section of a track, as shown in Figure. It rolls around the inside of a vertical circular loop 90.0 cm in diameter, and finally leaves the track at a point 20.0 cm below the horizontal section. (a) Find the speed of the ball at the top of the loop. Demonstrate that it will not fall from the track. (b) Find its speed as it leaves the track. (c) Suppose that static friction between ball and track were negligible, so that the ball slid instead of rolling. Would its speed then be higher, lower, or the same at the top of the loop? Explain.