EXAMPLE: Accelerometer A car has a constant acceleration of.0 m/s. A small ball of mass m = 0.50 kg attached to a string hangs from the ceiling. Find the angle θ between the string and the vertical direction. θ a
Free-body diagram: T B,S Net force must point to the right: F net on B W B,E T B,S a θ W B,E
In components: x: T x = ma Tsinθ = mg y: T y - W = 0 Tcosθ - mg = 0 y T a T cosθ = T y x θ T x = T sinθ W
equations, unknowns Tsinθ = ma Tcosθ - mg = 0 a mg tan T m a g g cos For a = m/s Note: θ does not depend on, θ = 1 the mass of the ball! Check: For a = 0 (constant speed), θ = 0
Lecture 7 Dynamics of Circular Motion
Circular motion The period (T ) of uniform (constant speed) cirular motion is the time needed by the rotating object to complete 1 full rotation. The frequency (f ) of uniform (constant speed) cirular motion is the number of rotations made by the rotating object in 1 unit time. f = 1 T The unit of frequency is hertz (Hz) 1 Hz = 1/s v = π r T
Dynamics of circular motion Conceptually, there s nothing new. All we need is an appropriate net force to produce the appropriate acceleration. To have circular motion, we always need a centripetal acceleration that is directed toward the center of trajectory, and Has the magnitude v a r We will always need a force: directed toward the center with magnitude F net = m v r Be careful: This is not some additional force! It is the net radial force. ( called centripetal force )
Who exerts the centripetal force?
EXAMPLE: Merry-go-round Little Jacob (15 kg) sits on the edge of a merry-go-round of radius 1.0 m while big sister makes it turn faster and faster. How fast can the system go before Jacob takes off if the coefficient of static friction between Jacob s pants and the merry-go-round is 0.5?
Little Jacob (15 kg) sits on the edge of a merry-go-round of radius 1.0 m while big sister makes it turn faster and faster. How fast can the system go before Jacob takes off if the coefficient of static friction between Jacob s pants and the merry-go-round is 0.5? N DEMO: Not enough f s static friction mg Static friction provides the needed radial acceleration: fs mr Maximum speed Maximum static friction: MAX S g r = (0.5) (9.8 m/s ) 1 m Smg mr =. rad/s 1 turn/s 3 MAX
Example: Curve A car of mass m with constant speed v drives through a curve of radius R. What is the minimum value of the coefficient of static friction between the tires and the road for the car not to slip? Fnet ma f s v m R f s a v m v R μ s mg f s μ s N = μ s mg μ s v gr
Example: Bucket A stone of mass m sits at the bottom of a bucket. A string is attached to the bucket and the whole thing is made to move in circles. What is the minimum speed that the bucket needs to have at the highest point of the trajectory in order to keep the stone inside the bucket? R
Example: Bucket A stone of mass m sits at the bottom of a bucket. A string is attached to the bucket and the whole thing is made to move in circles. What is the minimum speed that the bucket needs to have at the highest point of the trajectory in order to keep the stone inside the bucket? R N mg mg N ma a needs to be v a R v mg N m R
R N mg v mg N m R If v increases, N needs to be larger (if v becomes too large, since N is also the force on the bucket by the stone, the bottom of the bucket might end up broken ) If v decreases, N needs to be smaller. But at some point, N will become zero! This is the condition for the minimum speed at: v min mg = m R v min = gr The speed cannot get any smaller or the trajectory will not be a circle anymore (because the remaining forces mg-- will produce an acceleration that is too strong for a circle of radius R at that speed) DEMO: Bucket with water R If v < v min, the stone will do something like this...
Example: Car rounding a Banked Curve For a car moving around a banked curve it is possible to move at some speed when no friction is required. Find the banking angle if the speed is 15 m/s and the radius of a curve is 50 m. θ
Example: Car rounding a Banked Curve For a car moving around a banked curve it is possible to move at some speed when no friction is required. Find the banking angle if the speed is 15 m/s and the radius of a curve is 50 m. N cosθ N y x F x =ma x F y =ma y N sin θ = m v N cos θ mg = 0 R mg N sinθ θ tan θ = v gr = θ = 5 (15 m) (9.80 m/s ) (50 m) = 0.46
Example: Car on a bump The pavement on Grand Ave. is higher along the center of the street than along the sides. So when you drive along 13 th street and across Grand Ave., your car goes over a small hill. We can estimate the bump to have a radius of curvature of 30 m. What is the maximum speed your car should have if your wheels are to stay in contact with the ground all the time? A. 1 m/s B. 17 m/s C. 0 m/s R
Example: Car on a bump The pavement on Grand Ave. is higher along the center of the street than along the sides. So when you drive along 13 th street and across Grand Ave., your car goes over a small hill. We can estimate the bump to have a radius of curvature of 30 m. What is the maximum speed your car should have if your wheels are to stay in contact with the ground all the time? A. 1 m/s B. 17 m/s C. 0 m/s R
W N = ma y = v m R Large v -> Small N Small v -> Large N Smallest N : N =0 vmax Rg mg = = 17.1 m/s 60 mi/h v MAX m R N W R
EXAMPLE: Accelerometer A car has a constant acceleration of.0 m/s. A small ball of mass m = 0.50 kg attached to a string hangs from the ceiling. Find the angle θ between the string and the vertical direction. θ a
Free-body diagram: a θ
Free-body diagram: T B,S a θ W B,E
Free-body diagram: T B,S Net force must point to the right: F net on B W B,E T B,S a θ W B,E
In components: T a θ W
In components: y T θ a x W
In components: y T a T cosθ = T y x θ T x = T sinθ W
In components: x: T x = ma Tsinθ = mg y T a T cosθ = T y x θ T x = T sinθ W
In components: x: T x = ma Tsinθ = mg y: T y - W = 0 Tcosθ - mg = 0 y T a T cosθ = T y x θ T x = T sinθ W
Tsinθ = ma Tcosθ - mg = 0
equations, unknowns Tsinθ = ma Tcosθ - mg = 0
equations, unknowns Tsinθ = ma Tcosθ - mg = 0 tan a g
equations, unknowns Tsinθ = ma Tcosθ - mg = 0 a mg tan T m a g g cos
equations, unknowns Tsinθ = ma Tcosθ - mg = 0 a mg tan T m a g g cos For a = m/s, θ = 1 Check: For a = 0 (constant speed), θ = 0
equations, unknowns Tsinθ = ma Tcosθ - mg = 0 a mg tan T m a g g cos For a = m/s Note: θ does not depend on, θ = 1 the mass of the ball! Check: For a = 0 (constant speed), θ = 0