Physics Honors – Dr. Naumoff – Final Review, Torque and Static Equilibrium Torque=τ = Force applied x lever arm=F x r Torque – force not perpendicular to lever arm τ = F x r x sinθ Conditions for equilibrium The sum of all the forces (net force) acting on an object is zero. Fi 0 i The sum of all the torques (net torque) acting on an object is zero. i 0 i Total Clockwise Torques = Total Counterclockwise Torques (about a chosen axis). 1. A force of magnitude F =20.0 N is applied at a distance = 2.0m from the axis of rotation in an orientation where r makes the angle θ =90o with respect to the line of action of the force. What is magnitude of the torque at the axis of rotation? Fr 20.0 2.0 40.Nm 2. A force of 15.0 N is applied to a 50.0cm wrench at an angle of 45o to the lever arm. What is magnitude of the torque at the axis of rotation? Frsin 15.0 0.50 sin45o 5.3Nm 3. m1=1.0kg L1=1.0m; m2=1.0kg L2=0.5m; m3=3.0kg L3=?; mass of board =5.0kg. The fulcrum is supporting the board at its center. a. What is L3? cc 9.81 11 1 0.5 14.7 Nm c m3 g L3 L3 c m3 g 14.7 0.50m 3.0 9.81 b. What is the total force exerted upward by the fulcrum? Fup Fdown Fup 9.811 1 3 5 98.1N 4. Use the picture above for the following problems. m1=1.0kg; L=1.50m; mass of board =5.0kg ; x=.50m What is F1? Choose the left end of the board as the axis of rotation. cc c c 9.81 1.0 0.50 5.0 0.75 41.69 Nm cc cc 41.69 27.8 N 1.5 1.5 F1 Fdown F2 9.811.0 5.0 27.8 31.1N cc 1.5 F2 F2 5. A board is supported at each end. The board is 3.0m long and has a weight of 20.0N. There is a weight of 40.0N at a distance of 1.0m from the left end of the board. A weight of 60.0N is located at a distance of 2.00 m from the left end of the board. What are the upward forces at the ends of the board? Choose the left end of the board as the axis of rotation. cc c c 40.0 1.0 20.0 1.5 60.0 2 190 Nm cc Fup Fdown cc 190 63.3N L 3.0 F1 F2 40 20 60 120 N cc F2 L F2 F1 Fdown F2 120 63.3 56.7 N Dr. Naumoff Physics Honors – Final review Rotational kinematics and dynamics. 1 revolution 360o 2 radians Angular Displacement f f i t t f ti f i t t f ti For a rolling object vtranslational r =I net torque i o t For an object rotating in place K= o t 2 o2 2 1 2 (o )t For a rolling object 1 2 I 2 K= I moment of inertia 1 2 t 2 1 2 1 2 1 2 1 v mv I = mv I 2 2 2 2 r 2 angular acceleration the angular momentum of a rotating object is represented by the letter L kg m2 L=I unit; I moment of inertia angular velocity s angular momentum is conserved L=L W W=τ Δθ unit; N m=J power(P) P= W=Pt t 1) A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If the merrygo-round makes 4.9 rev/min, what is the velocity of the child in m/s? m 18m 4.9rev 2 rad 1min v r 4.62 s 2 1min 1rev 60s 2) Through what angle in degrees does a 33 rpm record turn in 1.0 s? o 33rot 360 1min o t 1s 198 1min 1rot 60s 3) A wheel accelerates from rest to 10 rad/s at a rate of 15 rad/s2. Through what angle (in radians) did the wheel turn while accelerating? 2 2 rad rad 2 2 10 0 s s 2 2 o o 2 3.33rad 2 rad 2 15 2 s 4) A wheel is accelerating from rest at a constant angular acceleration of 1.0 rad . s2 a. What is ϴ at t=1.0s? 1 1 2 t 2 0 11 0.50rad 2 2 b. What is ω at 1.0s? rad o t 0 11 1 s c. If the wheel has a moment of inertia I 2.0kg m2 .What is the torque acting on the ot wheel? I 2.01.0 2.0 Nm d. If the wheel has a diameter of 2.0 m, what is the tangential velocity of a point on its rim at t=3.0s? o t 0 1 3 3 rad s m 2 vtangential r 3 3 s 2 4) A hoop with a mass of 2.0 kg, and a radius of 0 .10m is rotating about an axis through its center at 10.0 rad 2 . The moment of inertia for a hoop is I mr . s a. Calculate the hoop’s moment of inertia. I mr 2 2.0 0.10 0.02kg m 2 2 b. Calculate the hoop’s kinetic energy. K 1 2 1 2 I 0.02 10.0 1.0 J 2 2 5) A hoop with a mass of 10.0 kg, and a radius of 0 .10m is rotating about the axis through its center at 15.0 rad 2 . The moment of inertia for a hoop is I mr . s a. Calculate the hoop’s moment of inertia. I mr 2 10.0 0.10 0.10kg m 2 2 b. Calculate the hoop’s kinetic energy. K 1 2 1 2 I 0.10 15.0 11.25 J 2 2 c. If the hoop were rolling on the ground at the above angular velocity, what would be its translational velocity? vtangential vtranslational r 0.10 15 1.5 m s d. If the hoop were rolling on the ground at the above angular velocity, what would be its kinetic energy? 2 1 1 1 2 1 v K= mv 2 I 2 = mv I 2 2 2 2 r 1 1 2 2 K= 10 1.5 0.10 15 22.5 J 2 2 1. A wave has a period of 2.50 s and a wavelength of 3.00 m. a. What is its speed? v f 0.40 3.0 1.2 m s b. What is its frequency? f 1 1 0.40 Hz T 2.5 c. The above wave transits to a different medium. Its frequency is unchanged, but its wavelength becomes 4.0 m. What is its new speed? v f 0.40 4.0 1.6 m s 2. Waves on a vibrating string are moving at 10.0 m . The frequency of vibration is s 15 Hz. The tension in the string is 100.0N. What is the string’s v F mass ratio ? length mass F F 100 kg v2 2 1.0 2 length v m 10 3. A closed tube is 2.50 m long. Assume that the speed of sound is 343m/s. a. What is the lowest frequency at which the column of air in this pipe will vibrate? 1 4 L 4 2.5 10m f1 v 343 34.3Hz 10 1 b. What is the next longest closed pipe that will vibrate at this frequency? 7.5m c. What is the minimal length of an open pipe that will vibrate at this frequency? 5.0m 4. (10 points) A steel string is 1.50m long and undergoing a tension of 500.0N. The value of µ for the string is 1.5x10-2 kg . m a. What is the value of the fundamental frequency of this string? F v = 500.0 m 182.6 2 1.5 10 s 1 2 L 2 1.5 3.0m f1 v 1 182.6 60.9 Hz 3.0 b. What are the values of the next three allowable harmonics? 121.8Hz; 182.7Hz;243.6Hz 5. Sketch the first three allowable harmonics for a 1.0m closed pipe. Label lengths, nodes, and antinodes. 6. In a different universe, light in a vacuum travels at a different speed than in our universe. Scientists have set up a Michelson experiment to measure the speed of light using an 10 sided rotating mirror and two fixed mirrors 30 000 meters from the rotating mirror. A clear image is obtained when the mirror is rotating at 200.0 rps (rotations per second). a. What is the speed of light in that universe? Let f the number of rotations per second or frequency 1 1 then the time required for one rotation is 5.0 10-3 s f 200 Time required for the mirror to advance one tenth of a rotation 1 5.0 10-3 s t 5.0 10-4 s rps number of faces 10 t is also the time required for light to make a round trip between the mirrors 2d 2 30 000 m 1.2 x108 -4 t 5.0 10 s s b. If the mirrors were 15 000 meters apart, how many rotations per second would the revolving mirror have to make to obtain a clear image? 1 t 5 104 s 2.5 104 s 2 1 1 rps 400rps number of faces t 10 2.5 104 s therefore c