Learningobject1
- 1. Neha Learning Object 2: Simple Pendulum Challenge Bill has a toy basketball with length 1.55m and a mass of 125g hanging off the wall of his room. Show your work: Q1: Find the net force at point P1 if the tension acting on the string is 0.5N. Q2: Billy swings the ball. What is the tension at point P2? (∅ = 43) Q3: What is the restoring force at P2 pulling back the ball to equilibrium? Q4: Calculate the acceleration of the tangential component at P2. Q5: Find the angular frequency of this oscillation. Q6: What is the displacement at point P2? Conceptual Questions: Q7: If the length of the string is increased to 3.0m, would the angular frequency increase or decrease? Q8: If the mass of the string is increased to 200g, would the angular frequency increase or decrease?
- 2. Bonus: What is the angle at ∅2? Answers: Q1: Since this is a mass hanging off of the string we have two forces to take into account, Fg and Ft. Solving for Fg: Fg=mg (125g/1000)(9.80m/s^2) =1.23N Solving for Fnet: Fnet=Fg-Ft Fnet=1.2325N-0.5N Fnet=0.7N Q2: Since the mass is now at an angle, we can solve for x and y components of forces for that angle. The tension in the string is equal to the force acting in the y direction (for every force there is an equal and opposite force!). Solving for T: T=mgCos∅ T=(.125kg)(9.80m/s^2)(Cos43) T=0.90N Q3: It is the tangential component that provides the restoring force that brings the ball back to equilibrium. Therefore, we must find the force acting in the x direction. Solving for Fr:
- 3. Fr=mgSin∅ Fr=(.125kg)(9.80m/s^2)(Sin43) Fr=0.84N Since this force is in the opposite direction, Fr=-0.84N Q4: If we have the length of the string, we can use the angular frequency formula. W(angular frequency)=√( 𝑔 𝐿 ) W=√ 9.80𝑚/𝑠^2 1.55𝑚 W=2.51 Hz Q6: The displacement at P2 depends on the length of the string and the angle at which the ball is at. 180 degrees= 𝜋 radians therefore, 43 degrees=0.750 radians S=L∅ S=(1.55m)(0.750 radians) S=1.16m Q7: Since the equation of angular frequency is√( 𝑔 𝐿 ) , increasing the length of the string will decrease the angular frequency. Q8: There is no change. Angular frequency does not rely on mass, but on length!
- 4. Bonus: At P1, the angle of the ball with respect to the wall is 90 degrees. If ∅1=43 degrees, then: ∅2=90-43=47 degrees