Question

A simple pendulum that has a length of 1.7 m and a bob with a mass of 4.0 kg is pulled an angle θ from its nătural cquilibrium position (hanging straight down) and then let go. Define the zero point of the bob's gravitational potential energy at the fowest point of its motion. For θ=5.0 *, what are the maximum gravitational potential energy Umax(5 ⁰) and maximum speed vmax(5 ⁰) of the bob? Umin(5 *)= vmax(5*)= m/s For θ=10∘, what are the maximum gravitational potential energy Umax(10∘) and maximum speed Umax(10 ⁰) of the bob? For θ=15 *, what are the maximum gravitational potential energy Umax(15∗) and maximum speed vman(15*) of the bob? Umax(15′)= For θ=20 ⁰, what are the maximum gravitational potential energy Umax(20⁰) and maximum speed vmax(20 ⁰) of the bob? Umax(20 ⁰)= J vmax(20 ⁰)= For θ=25∗, what are the maximum gravitational potential energy Umax(25 *) and maximum speed vmax(25 *) of the bob? Umax(25 *)=

Answer 1

To find the maximum gravitational potential energy and maximum speed of the bob at different angles, we can use formulas that involve the length of the pendulum and the angle θ.

Step-by-step explanation:

To find the maximum gravitational potential energy (Umax) of the bob at an angle θ, we can use the formula Umax = mgh, where m is the mass of the bob, g is the acceleration due to gravity, and h is the height of the bob above its lowest point. Since the bob is pulled from the natural equilibrium position, the height h can be determined as L(1-cosθ), where L is the length of the pendulum.

To find the maximum speed (vmax) of the bob at angle θ, we can use the formula vmax = √(2gh), where g is the acceleration due to gravity and h is the height of the bob above its lowest point. The height h can be determined as L(1-cosθ), where L is the length of the pendulum.