  42. A pendulum with a cord oflength r 1.00 m swings ina vertical plane (Fig. P4.42)When the pendulum is in thetwo horizontal positions 090.0° and 0 270°, its speed is5.00 m/s. Find the magnitudeof (a) the radial accelerationand (b) the tangential acceleration for these positions(c) Draw vector diagrams to=determine the direction of thetotal acceleration for these twoposiions. (d) Calculate themagnitude and direction ofthe total acceleration at theseFigure P4.42two positions

Question help_outlineImage Transcriptionclose42. A pendulum with a cord of length r 1.00 m swings in a vertical plane (Fig. P4.42) When the pendulum is in the two horizontal positions 0 90.0° and 0 270°, its speed is 5.00 m/s. Find the magnitude of (a) the radial acceleration and (b) the tangential accel eration for these positions (c) Draw vector diagrams to = determine the direction of the total acceleration for these two posiions. (d) Calculate the magnitude and direction of the total acceleration at these Figure P4.42 two positions fullscreen
Step 1

Part A: help_outlineImage TranscriptioncloseThe cord length of the pendulum is 1.00 m swings in a vertical plane and speed is 5.00 m/s. Write the formula to calculate the radial acceleration Here, v is the speed of cord, r is the length of the cord and a is the radial acceleration of the cord fullscreen
Step 2 help_outlineImage TranscriptioncloseSubstitute 5.00 m/s for v and 1m for r to find a. (5.00 m/s)2 a 1m The radial acceleration is same for both condition at 0=90° and 0 = 270* Conclusion Therefore, the radial accelerations at the both are 25.0 m/s2. position 90 and 270 fullscreen
Step 3

Part B:... help_outlineImage TranscriptioncloseFor the pendulum, the tangential acceleration is only due to that of gravity and it is downward at both position Conclusion: Therefore, the tangential acceleration at and 270 both the position 90 9.81 m/s2 is fullscreen

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