Question

The balance wheel of a watch oscillates with angular amplitude 1.2π rad and period 0.70 s. Find (a) the maximum angular speed of the wheel, (b) the angular speed of the wheel at displacement 1.2π/2 rad, and (c) the magnitude of the angular acceleration at displacement 1.2π/4 rad.

Answer 1

a )

given

angular amplitude 1.2 π rad and period 0.70 s

displacement of the wheel θ = θ^' cos( 2 π t / T )   

angular velocity ω = - ( 2 π / T ) θ^' [ sin( 2 π t / T ) ]

ω_max = ( 2 π / T ) θ^'  

= ( 2 π / T ) ( 1.2 π )

= ( 2 π / 0.7 ) ( 1.2 π )

ω_max = 33.80 rad/sec

b )

the angular displacement θ = 1.2 π / 2

θ = θ^' cos( 2 π t / T )

θ / θ' = cos( 2 π t / T )

cos( 2 π t / T ) = θ / θ'

= [ 1.2 π / 2 ] / [ 1.2 π ]

cos( 2 π t / T ) = 1 / 2

sin( 2 π t / T ) = [1 - cos²( 2 π t / T ) ]^1/2

= [ 1 - ( 1 / 2 )² ]^1/2

sin( 2 π t / T ) = ( √3 ) /

angular velocity ω = - ( 2 π / T ) θ^' [ sin( 2 π t / T ) ]

= - ( 2 π / 0.7 ) ( 1.2 π ) [ ( √3) / 2 ]

ω = - 29.275 rad/sec ( for another cycle is ω = 29.275 rad/sec )

c )

angular acceleration is α = dω / dt

= - ( 2 π / T )² θ^' [ cos( 2 π t / T ) ]

given angular displacement θ = (1.2) ( π / 4 )

θ = θ^' cos( 2 π t / T )    

θ / θ' = cos( 2 π t / T )

cos( 2 π t / T ) = θ /θ'

= [ ( 1.2 )( π / 4)] / [ 1.2 π ]

cos( 2 π t / T ) = 1/4

α = - ( 2 π / T )² θ^' [ cos( 2 π t / T ) ]

= - ( 2π / 0.7 )² ( 1.2 π ) [ 1 / 4 ]

α = - 75.81 rad/s² or ( α = 75.81 rad/s² )