A horizontal bar 1.5 metres long and of small cross-section rotates about vertical axis through one end. It accelerates uniformly from 1200 r. P.M. To 1500 r. P.M. In an interval of 5 seconds. What is the linear velocity at the beginning and end of the interval ? What are the normal and tangential components of the acceleration of the mid-point of the bar after 5 seconds after the acceleration begins ?

a. i. 188.49 m/s. ii. 235.62 m/s. b. i. 117.81 m/s² ii. 4.71 m/s²

Explanation:

a. Since the linear velocity v = rω where r = radius of path = length of horizontal bar = 1.5 m and the initial angular speed ω₀ = 1200 r.P.M = 1200 × 2π/60 = 125.66 rad/s and final angular speed after the 5 seconds ω₁ = 1500 r.P.M = 1500 × 2π/60 = 157.08 rad/s

i. So, the initial linear velocity v₀ = rω₀ = 1.5 m × 125.66 rad/s = 188.49 m/s.

ii. The final linear velocity v₁ = rω₁ = 1.5 m × 157.08 rad/s = 235.62 m/s.

b. The mid-point of the bar is at a distance of one-half from the end of the bar. So its radius r' = r/2 = 1.5 m/2 = 0.75 m

i. Its normal acceleration a after the 5 seconds is a = r'ω₁² = 0.75 m × 157.08 rad/s = 117.81 m/s²

ii. Its tangential acceleration is given by a' = r'α where α = tangential acceleration

α = (ω₁ - ω₀)/t  where t = 5 seconds

α = (157.08 rad/s - 125.66 rad/s)/5 s

= 31.42 rad/s ÷ 5 s

= 6.28 rad/s²

a' = r'α

= 0.75 m × 6.28 rad/s²

= 4.71 m/s²

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!