Question

003 10.0 points Earth orbits the sun once every 365.25 days. Find the average angular speed of Earth about the sun. Answer in units of rad/s. 004 (part 1 of 2) 10.0 points A disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and magnitude of the acceleration of a bug clinging to the rim of the disk? 1. 1 m/s and 0 m/s2 (Disk spins at constant speed.) 2. 10 m/s and 10 m/s2 3. 1 m/s and 10 m/s2 4. 0.1 m/s and 1 m/s

Answer 1

Solution) (3) Time T = 365.25 days

Here we have time T in days so we should convert it into seconds as shown below

T = 365.25 days

1 day = 24 hours (1 hour =60minutes)

1 day =24 × 60 minutes ( 1 minute =60 seconds)

1 day =24×60×60 seconds

T = 365.25 × 24×60×60 seconds

T = 31557600 seconds

Therefore average angular speed w = (2(pi))/(T)

w = ( 2×3.14)/(31557600)

w = 0.000000199 rad/s

w =0.199×10^(-6) rad/s

(4) radius r =0.1m

Angular speed w=10rad/s

Speed V = rw = 0.1×10=1m/s

V= 1m/s

Acceleration a = (V^2)/(r)

a = (1^2)/(0.1)

a= 10 m/s^2

Therefore V= 1m/s and a =10m/s^2

Option (3) is correct