Question

A rigid link PQ of length 1.0 m is pinned at P. It rotates about P in a vertical plane with a uniform angular acceleration of 1.0 rad/s² . At an instant when the angular velocity of the link is 1.0 rad/s, the magnitude of total acceleration (in m/s² ) of point Q relative to point P is

1. 1.41
2. 1.73
3. 2
4. 2.83

Answer 1

Correct Answer - Option 1 : 1.41

Concept:

Tangential acceleration (at):

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where, α = angular acceleration and r = radius

at = rα

Centripital acceleration (ac):

ac = ω2r

where,  ω = Angular Velocity

Calculation:

Given:

α = Angular acceleration = 1 rad/s2

r = radius of crank = 1 m

ω = Angular Velocity = 1 rad/s

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

∴ at = rα = 1 × 1 = 1 m/s2 

∴  ac = ω2r = (1)2 × 1 = 1 m/s2 

\(\therefore\ \text{Total acceleration a} = \sqrt {a_t^2 + a_c^2} = \sqrt {{1^2} + {1^2}} = \sqrt 2 =1.41 \;{m}/{{{s^2}}}\)