This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Okay, Divine, let's tackle this new question 2. 2. Circular Motion a) The frequency of the motion Step 1: Identify the given values for the number of revolutions and time. Number of revolutions (N) = 150 Time (t) = 5 minutes Step 2: Convert time to seconds. t = 5 minutes × 60 s/minute = 300 s Step 3: Calculate the frequency (f) using the formula f = (N)/(t). f = 150 revolutions300 s f = 0.5 Hz The frequency of the motion is 0.5 Hz. b) The centripetal force on the stone Step 1: Identify the given values and convert them to SI units. Mass (m) = 100 g = 0.1 kg Diameter (D) = 80.0 cm = 0.80 m Radius (r) = (D)/(2) = 0.80 m2 = 0.40 m Frequency (f) = 0.5 Hz (from part a) Step 2: Calculate the angular velocity () using the frequency. = 2 f = 2 (0.5 Hz) = rad/s Step 3: Calculate the centripetal force (F_c) using the formula F_c = m^2 r. F_c = (0.1 kg)( rad/s)^2 (0.40 m) F_c = (0.1)(0.4)^2 N F_c = 0.04^2 N Using ≈ 3.14159: F_c ≈ 0.04 × (3.14159)^2 N F_c ≈ 0.04 × 9.8696 N F_c ≈ 0.395 N The centripetal force on the stone is 0.04^2 N or approximately 0.395 N. c) Sketch a distance-time graph for this motion of the shadow for two cycles The shadow of the stone moving in a horizontal circle undergoes Simple Harmonic Motion (SHM) on the wall. • The amplitude of this motion is the radius of the circle, A = r = 0.40 m. • The period of the motion is T = (1)/(f) = (1)/(0.5 Hz) = 2 s. • The graph will be sinusoidal, showing the shadow's distance from the center of its oscillation over time. For two cycles, the graph will span 2T = 4 s. Here is a description of the sketch: The graph will have "Time (s)" on the x-axis and "Distance from center (m)" on the y-axis. The y-axis will range from -0.40 m to +0.40 m. The x-axis will range from 0 s to 4 s. The curve will be a sine wave (assuming the shadow starts at the center of its oscillation at t=0) or a cosine wave (if it starts at maximum displacement). It will complete one full oscillation (return to its starting position and direction) every 2 s. So, at t=0, 2, 4 s, the distance will be 0 m (if starting from center). At t=0.5, 2.5 s, the distance will be +0.40 m. At t=1.5, 3.5 s, the distance will be -0.40 m. 3 done, 2 left today. You're making progress.