Motion Part 4: Projectile Motion Last modified: 28/03/2017
CONTENTS Projectile Motion Uniform Motion Equations Projectile Motion Equations Trajectory How to Approach Problems Example 1 Example 2 Example 3 Rotational Motion Angular Speed Angular Velocity: Definition Example 1 Tangential Velocity Examples 2 & 3 Acceleration Centripetal Acceleration Example 4 Constant Angular Acceleration Example 5 Summary
Uniform Motion Equations In the last lecture we saw the equations of uniform motion (i.e. where acceleration a is constant), which connect the time t, displacement s, initial and final velocities u and v: v = u + at s = ut + 1 2 at2 v 2 = u 2 + 2a s In that lecture we looked at applying these formulas to linear motion (i.e. motion in one dimension only). The next type of problem we will look at, is that of projectile motion - motion in two dimensions with the acceleration due to gravity.
Projectile Motion Equations We have already considered the cases of an object being thrown vertically up, or dropped down. The more general case has the initial velocity u at an angle to the acceleration a: y time t v a = gj u s t = 0 x The object follows a curved path and at time t, it has displacement s and velocity v.
We can use our general equations to find expressions for the three basic quantities a, v and s: a = gj v = u gtj s = ut 1 2 gt2 j As we will see, these equations are used similarly to the linear motion case. But first we will determine the shape of the trajectory of the object. The trajectory is the path the object follows through space.
Trajectory The mathematical equation describing the trajectory will express the y co-ordinate of the position at time t in terms of the x co-ordinate at the same time. To do this, we write the initial velocity u and displacement s in terms of their components: u = u x i + u y j and s = xi + yj Thus: s = ut 1 2 gt2 j xi + yj = u x ti + (u y t 1 2 gt2 )j Comparing the two sides of this equation gives: x = u x t and y = u y t 1 2 gt2
The first equation gives t = x/u x which we can use to eliminate t from the second equation: y = u y ( x u x ) 1 2 g ( x u x ) 2 = ( uy u x ) ( ) g x 2ux 2 x 2 Remembering that g, u x and u y are all constant, this is the equation of a parabola. Projectiles acting only under gravity always follow a parabolic path.
How to Approach Problems Projectile problems are similar to linear motion problems, but can be little more complicated. Read the problem carefully, and make a list of the information you are given. Draw a diagram, and pay particular attention to getting the correct directions. Remember, there is now motion in two dimensions - it is usually best to separate the x and y variables: x motion a x = 0 y motion a y = g u x = v x u y v y s x t s y t There are six different variables involved - ux, u y, s x, s y, v y and t The speed in the x direction is constant The time t is the same for both x and y directions.
From here the calculation usually requires two steps : (i) Use the information about one direction to calculate the time t using the one-dimensional equations. (ii) Use this calculated value together with the information about the other direction to calculate the required quantity. Because of this two-step process and the large of number of variables you need to keep track of, it is very easy to get mixed-up. It is very important to be organized and methodical in these calculations.
Example 1 A rock is thrown into the air at 50 m/s in a direction 37 above the horizontal. Calculate: (a) the maximum height reached by the rock (b) the horizontal distance R to the point of impact with the ground. ( the range of the rock) First draw a diagram and collect the known information: v y = 0 a y = g = 10 m/s 2 u = 50 m/s uy = 30 m/s h max =? t = t flight ux = 40 m/s R = s x at t = t flight
(a) Calculating the maximum height only requires information about the y direction: v 2 y = u 2 y + 2a y s y 0 = 30 2 + 2( 10)s y s y = 900 20 = 45 m This is exactly the same calculation seen in Example 3a in the last lecture. For this particular calculation the x motion is irrelevant. (b) The range R does involve the x direction: R = u x t flight = 40 t flight but we have insufficient information to solve this using only the x motion.
The time of flight must be calculated using the y variables (in exactly the same way as Example 3b in the last lecture), using the fact that s y = 0 when t = t flight. s y = u y t + 1 2 a yt 2 0 = 30t + 1 2 ( 10)t2 0 = 6t t 2 0 = t(6 t) t = 0 s or t = 6 s Again, t = 6 s is the required solution. (We could also calculate this as twice the time to reach maximum height) Using this value in the previous expression gives the answer: R = u x t flight = 40 6 = 240 m
Example 2 A ball is thrown into the air at 50 m/s in a direction 37 above the horizontal from the edge of a 35 m cliff. Calculate the horizontal distance R to the point of impact with the ground. Again a diagram helps to collect the given information: a y = g = 10 m/s 2 u = 50 m/s uy = 50 sin 37 = 30 m/s 37 ux = 50 cos 37 = 40 m/s 35 m t = t flight R = s x at t = t flight
The method here is exactly the same as we have just used. To calculate the range we need to first determine the time of flight, using the y direction only. This time of flight will be the time when the displacement in the y direction is s y = 35 m. s y = u y t + 1 2 a yt 2 35 = 30t + 1 2 ( 10)t2 0 = 7 + 6t t 2 0 = (t 7)(t + 1) t = 7 s or t = 1 s The first of these is clearly the correct one in this case. (Note that this calculation is exactly the same as done in Example 4a from the previous lecture) We now use t flight = 7 s in the x direction equation: R = u x t flight = 40 7 = 280 m
Example 3 A bottle is located on top of a 10 m high vertical pole. A ball is thrown from a point on the ground 20 m from the base of the pole, at an angle θ = sin 1 4 5. What is the initial speed of the ball, u, if the ball hits the bottle? This problem has two unknowns: initial speed u and the time of flight, T. a y = g = 10 m/s 2 t flight = T 10 m u =? uy = 4 5 u ux = 3 5 u 20 m
To solve for the two variables u and T, we need two equations: From the motion in the y-direction: s y = u y t + 1 2 a yt 2 10 = 4 5 ut + 1 2 ( 10)T 2 and from the motion in the x-direction: s x = u x t 20 = 3 5 ut Eliminating T from these equations gives (after a bit of algebra) our required solution: u = 18.3 m/s
Motion Part 5: Rotational Motion
Angular Speed A circular turntable is rotating about its axis as shown below. Fluffy the cat jumps on to the turntable and sits to enjoy the ride at a point located a distance r from the axis of rotation. rotation How can we best describe Fluffy s motion? r Using velocity is complicated, because the direction of the motion is constantly changing. Let s try to keep it simple: What is Fluffy s speed?
If the time taken for one revolution is T, then the average speed of Fluffy will be the distance travelled in one revolution divided by T. distance = 2πr r v average = 2πr T = ( 2π T )r However, if Fluffy moves to a position with a different value of r, then this speed will also differ. Different points on the turntable have different speeds. It therefore doesn t make a lot of sense to refer to the speed of the turntable. What is constant at all points on the turntable is the term in red - the angle in radians divided by the time taken. This is known as the angular speed of the turntable (and Fluffy!).
Angular Velocity: Definition The proper definition of angular velocity is similar to that of regular velocity. Consider two points in a circular path: A is the position at time t and B is the position at t + t. r B B θ The average angular velocity is defined as A the change in angle (in RADIANS) over the change in time: r A ω average = θ t The instantaneous angular velocity is defined by taking the limit: θ ω = lim t 0 t = dθ dt [Note: ω is the Greek letter omega, NOT w ]
The SI unit for angular velocity is radians per second (rad/s) NOTE: the dimensions of angular velocity (T 1 ) are different to those of velocity (LT 1 ) It may not be obvious, but angular velocity is a vector! thumb gives direction of the vector ω The direction is given by a right hand rule. (Which indicates there is a vector product involved. We ll see how in a future lecture.) grab axis of rotation and curl fingers in direction of rotation The vector-ness of ω is only really important when we are dealing with complicated motions with multiple rotations occuring at once around different axes. We won t be seeing any such problems in this course.
Example 1 Vinyl records are usually played at speeds of 33 1 3 are these angular speeds in SI units? or 45 rpm. What Revolutions per Minute (rpm for short) is a commonly used non-si unit for angular velocity. If we need to calculate with these values, we must first convert to the SI unit: radians/second. This is simple if we realize: 1 revolution = 2π radians, and of course 1 minute = 60 seconds. Therefore: and 33 1 3 100 rpm = 3 2π = 3.5 rad/s 60 45 rpm = 45 2π = 4.7 rad/s 60
Tangential Velocity The position vector r of an object travelling in a circular path can be wriiten in terms of the unit vectors i and j, as: y r θ r cos θ r sin θ x r = r cos θi + r sin θj = r(cos θi + sin θj) = r ˆr where θ is a function of time. Remembering the definition of velocity, we have: v = dr dt = r d (cos θi + sin θj) dt = r dθ ( sin θi + cos θj) = r ω ˆv dt and we see that the speed is v = rω, as seen earlier for Fluffy.
What is the direction of v? Using the 2 expressions on the previous page: r v = {r(cos θi + sin θj)} {rω( sin θi + cos θj)} = r 2 ω ( cos θ sin θ + sin θ cos θ) = 0 The velocity vector v is perpendicular to the position vector r, i.e. the velocity vector is tangent to the circular path. For this reason v is known as the tangential velocity. In fact, for any motion, the instantaneous velocity vector is always tangent to the trajectory.
Examples 2 & 3 An ant is sitting at a distance of 12 cm from the centre of a turntable rotating at a constant 33 1 3 rpm. What is this ant s tangential speed? From earlier, we know: 33 1 3 rpm = 3.5 rad/s, so: v = rω = 0.12 3.5 = 0.42 m/s At what distance from the centre of the turntable would the ant s speed be 0.15 m/s? r = v ω = 0.15 3.5 = 0.043 m = 4.3 cm
Acceleration Next we can calculate the acceleration vector: a = dv dt = d [rω ( sin θi + cos θj)] dt = r dω dt ( sin θi + cos θj) + rω2 ( cos θi sin θj) = r dω dt ˆv rω2 ˆr = rα ˆv rω 2 ˆr where we introduce the angular acceleration α = dω dt (SI units: rad/s 2 ) Notice there are two components in the acceleration vector - the first is in the same direction as the velocity, and is known as the tangential acceleration a T = rα. This part of the acceleration changes the speed of the object. If the motion has a constant angular velocity ω, then this term will be ZERO.
Centripetal Acceleration The second component of the acceleration vector is the really important one: rω 2 ˆr The direction is opposite to r, i.e. towards the center of the circle. Called the centripetal acceleration a c, this is what is keeping the object in the circular path - it doesn t change speed, only the direction of motion. No centripetal acceleration No circular motion. Remembering that speed v = rω: a c = rω 2 = v 2 r
Example 4 For the ant at r = 12 cm on the 33 1 3 rpm turntable, what are the ant s (a) tangential acceleration, and (b) centripetal acceleration? (a) We are told the turntable is rotating at a constant angular speed. Constant = no change = 0 tangential acceleration. (b) We know both ω and v, so there are two equivalent ways to calculate a c : or a c = ω 2 r = (3.5) 2 0.12 = 1.46 m/s 2 a c = v 2 /r = (0.42) 2 /0.12 = 1.46 m/s 2
Constant Angular Acceleration Consider an object following a circular path of radius r and constant angular acceleration α, from t = 0 where the angular speed is ω 0 to a later time t, with angular speed ω. We can calculate the initial and final velocities, acceleration and distance travelled as shown at right. If we now imagine unrolling the circular path, we can apply our equations for linear motion with constant acceleration: v = rω r θ a T = rα s = rθ u = rω 0 v = u + at rω = rω 0 + rαt ω = ω 0 + αt s = ut + 1 2 at2 rθ = rω 0 t + 1 2 rαt2 θ = ω 0 t + 1 2 αt2 v 2 = u 2 + 2as (rω) 2 = (rω 0 ) 2 t + 2(rα)(rθ) ω 2 = ω 2 0 + 2αθ
We can apply these equations in the same way as we did in the linear motion examples seen previously. Note that the rotational equations are the same as the regular equations, but with the substitutions v ω, a α, s θ. This same idea will apply more generally, for example: force F = ma torque τ = Iα momentum p = mv angular momentum L = Iω kinetic energy KE = 1 2 mv 2 rotational kinetic energy KE = 1 2 Iω2... where the moment of inertia I is the rotational equivalent of mass. We will briefly use the first of these later in the course.
Example 5 An ant is sitting at a distance of 12 cm from the centre of a stationary turntable, which accelerates uniformly to an angular speed of 33 1 3 rpm in 0.5 s. (a) What is the ant s angular acceleration? (b) Through what angle does the ant move in this 0.5 s interval? (a) We know: ω 0 = 0, ω = 33 1 3 so: rpm = 3.5 rad/s and t = 0.5 s ω = ω 0 + αt α = (ω ω 0 )/t = 3.5 0 0.5 = 7.0 rad/s 2
(b) We already knew: ω 0 = 0, ω = 3.5 rad/s and t = 0.5 s. We now also know: α = 7.0 rad/s 2, so similarly to previous examples we have a choice of methods to determine the angular displacement θ: or θ = ω 0 t + 1 2 αt2 = 0 + 1 2 7.0 (0.5)2 = 0.875 rad ω 2 = ω 2 0 + 2αθ θ = ω2 ω 2 0 2α = 3.52 0 2 7.0 = 0.875 rad Of course, as we should expect, both methods give the same result.
Summary Different points on a rotating object will have different values of speed and acceleration. All points will have the same values of angular velocity and angular acceleration. Angular Speed: ω = dθ dt Tangential Speed: v = rω Angular Acceleration: α = dω dt Tangential Acceleration: a T = rα Remember the word tangential is describing direction. Tangential velocity is a proper velocity with dimensions LT 1 and units m/s. Angular velocity is a dimensionally different quantity, and is NOT a velocity. Similarly, while tangential acceleration is a proper acceleration, angular acceleration is dimensionally different.
To continue in a circular path, there must be a component of ( proper ) acceleration called the centripetal acceleration directed TOWARDS THE CENTRE OF THE CIRCULAR PATH. ( centripetal is again referring to the direction of the acceleration) The magnitude of the centripetal acceleration must have exactly the right value for the path to be circular: a c = ω 2 r = v 2 /r y v r θ r 0 a c x In the case of constant angular velocity, then α = 0 and a T = 0 but the centripetal acceleration must have the value given by the above formula, and be directed towards the centre of the circle.