The100-kg pendulum has a center of mass at G and a radius of gyration about G of kg = 250 mm. Determine the horizontal and vertical components of reaction on the beam by the pin A and the normal reaction of the roller B at the instant 6 = 07 when the pendulum isrotating at @ = 4 rad/s. Neglect the weight of the beam and the support.
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- The rod shown, released from rests in this position (0 = 0) and the spring is unstretched. If the body will stopped momently at e = 49 degrees, Find the work done by slender weight in joules. Given L=1.3 m, mass=20 Kg wwww B k 3L/4 L/4 Answer:The 30-kg uniform thin hollow square plate is pinned at point O, and its side L = 0.2 m. If it is subjected to the constant moment M = 62 N•m and is released from rest from the position as shown, determine the total work done to it when it has rotated 45. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point, and proper Sl unit. Take g = 9.81 m/s?. M Your Answer: Answer unitsThe 150-kg pendulum has a center of mass at G and a radius of gyration about G of kG = 0.25m, as shown in Figure 2. Determine the horizontal and vertical components of reaction on thebeam by the pin A and the normal reaction of the roller B at horizontal direction when thependulum is rotating at 5 rad/s. Neglect the weight of the beam and the support. If the normalreaction of the roller B at horizontal direction change to vertical direction, determine thehorizontal and vertical components of reaction on the beam by the pin A.
- Find the stable equilibrium position of the system described in Prob. 10.56 if m = 2.06 kg.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 ke=0.4 m. The spring's unstretched length is Lo=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 8-30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is 0=0°. The spring's length at the state 2 is L2=4 m. (4) The elastic potential energy at the potion 1 is_ HULKU 2₂ State 2 G m State 1 (N-m) (two decimal places)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 ke=0.4 m. The spring's unstretched length is Lo=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 8-30°. The wheel rolls without slipping and passes the position at the state 2 when the angle is 8=0°. The spring's length at the state 2 is L2=4 m. (6) The elastic potential energy the state 2 is HILAI L₂ # State 2 ZG State 1 (N-m) (two decimal places)
- 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. Ignore the spring's mass. (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) (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) (3) At state 1, how long the spring is stretched from its unstretched state (length difference):________(m) (two decimal places) (4) The…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 mass center G is set as the origin (datum), the gravitational potential energy at the state 1 is___ (two decimal places)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. (7) The instantaneous center of zero velocity (IC) is A. Point A B. Point O C. Point G
- 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. Ignore the spring's mass. (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___ N m(two decimal places) (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___ N m (two decimal places) (3) At state 1, how long the spring is stretched from its unstretched state (length difference):________(m) (two decimal places) (4) The…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. (4) The elastic potential energy at the potion 1 is_______(N·m) (two decimal places)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 mass center G is set as the origin (datum), the gravitational potential energy at the state 1 is___ (two decimal places) (2) If the mass center G is set as the origin (datum), the gravitational potential energy at the state 2 is___ (two decimal places) (3) The stretched spring length of the spring at the state 1 is________(m) (two decimal places) (4) The elastic potential energy at the potion 1 is_______(N·m) (two decimal places) (5) The stretched spring length of the spring at the…