TS EAMCET 4 May 2018 Shift 2 Solved Paper

Consider the following figure.As we know that,
v=ωR
ω=

v

R

Substitute the values.
ω=

2

5

rad∕s
And,
The expression for time period is given as,
T=

2π

ω

Substitute the values.
T=

2π

2 5

=5π for one revolutionCalculate T for 2 revolutions as follows,
T=5×2 ×π
=10πs
The expression of acceleration is given by,
α=

v2

R

=

22

5

=0.8ms−2