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

it's a b and

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j