A simple pendulum is made of a 50 cm-string and a bob of mass m. At t = 0, the pendulum is at its equilibrium position and is given an initial velocity v = 0.1 m/s. The maximum angular speed. 6'max, is: %3D 0.08 rad/s 02 rad/s 1005 rad/s. 04rad/s
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- A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.0400 m. A 0.200 kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm (Fig.). Find the maximum distance the frame moves downward from its initial equilibrium position.consider the mass and pulley system in the attached file. mass m1 = 29 kg and mass m2 = 12kg. the angle of the inclined plane is given and the coefficient of kinetic friction between mass m2 and the inclined plane is uk = 0.12. assume the pulleys are massless and frictionless if the system is released from rest, what is the speed of mass m2 after 2.5 s? v(2.5 s) = ?consider the mass and pulley system in the attached file. mass m1 = 29 kg and mass m2 = 12kg. the angle of the inclined plane is given and the coefficient of kinetic friction between mass m2 and the inclined plane is uk = 0.12. assume the pulleys are massless and frictionless when mass m2 moves a distance 4.94 m up the ramp, how far downward does mass m1 move? d= ?
- Determine the magnitude and direction of the angle theta of the force FAB that must be exerted on bar AB to maintain equilibrium of the system. The suspended mass is 110 kg. Neglect the size of the pulley at A and treat all pulleys as smooth. What should happen to maintain the equilibrium of the system if FAB acts with an angle theta= 0°?Consider g = 9.81 m/s2.While the system shown is in equilibrium, a counter-clockwise torque T =T0sin(ωt) N-m(with T0 = 226.6 and ω=7) is applied to teh disk. What is the disk angular speed ω0 = ? in rad/s at time t = 4.2 sec? ( I = 0.2 kg.m2, b = 7.8 N-m-s/r, k = 1,504 N/m, r = 0.52m )A 4 kg mass suspended from a spring having stiffness of k= 600N/m. if the block is pushed50mm upward from its equilibrium position and then released from rest dertermine the equation which describes the blocks motion
- Please answer: 3, 4, 5, 6, 8, 9, 10, 12 The wheel is attached to the spring. The mass of the wheel is m=20 kg. The radius of the wheel is 0.6m. The radius of gyration kG=0.4 m. The spring’s unstretched length is L0=1.0 m. The stiffness coefficient of the spring is k=2.0 N/m. The wheel is released from rest at the state 1 when the angle between the spring and the vertical direction is θ=30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is θ=0°. The spring’s length at the state 2 is L2=4 m. (1) If the datum for gravitational potential energy is set as shown below, the the gravitational potential energy of the wheel at the state 1 is 0 N m(two decimal places) (ANSWER IS 0) (2) If the datum for gravitional potential energ is set as shown below, the gravitational potential energy of the wheel at the state 2 is 0 N m (two decimal places) (ANSWER IS 0) (3) At state 1, how long the spring is stretched from its unstretched state (length…13. An object (at earth) at rest with a mass of 3kg is being dropped 3m above the spring whose constant is 10N/mm. If the spring has to stopped the falling object, determine the magnitude of deflection of the spring right after the stoppage. Assume no energy loss during operation. Derive with the formula below, show the cancellation, explain the step by step and DRAW a free body diagram/figure. refer with this: PE1 + KE1 + U1 + WF1 + Q1 + W1 = PE2 + KE2 + U2 + WF2 + Q2 + W2 + E losses.Dynamics Question If the normal acceleration of the rods is just w^2 * 1.5m, is the normal acceleration of the 50-kg bar just w^2* 2m?
- Please answer this NEATLY, COMPLETELY, and CORRECTLY for an UPVOTE. Block D, initially at rest, having a W = 30 lbf , is supported within the system shown, with x = 13.2 ft. Neglecting the mass of the pulley and the cord on the system, determine the eventual height h after a force P = 60 lbf is applied to make the block rise by ∆sD = 0.625 ft at t = 2.5 s. NOTE: Use Force-Mass-Acceleration (FMA) Method.mass = 20 kg At the end of 20 complete cycles, t = 3 s, amplitude = 0.5 cmConsider a slender rod AB with a length l and a mass m. The ends are connected to blocks of negligible mass sliding along horizontal and vertical tracks. If the rod is released with no initial velocity from a horizontal position as shown in Fig.A, determine its angular velocity after it has rotated through an angle of θ (see Fig B) using the conservation of energy method. (Hint: Moment of inertia of rod about G = (1/12)ml2 The kinetic energy of a rigid body in plane motion is