Question

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in the horizontal position θ = 0 where the spring is unstretched. If the bar is observed to momentarily stop in the position θ = 64°, determine the spring constant k. For your computed value of k, what is magnitude of the angular velocity of the bar when θ = 32°.

Answer 1

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Step-by-step explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

`T_1+V_1=T_2+V_2`

`0+0 = (1)/(2) k \delta^2 - (mg (a+b) sin \ \theta )/(2) \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = (mg(a+b) sin \ \theta )/(\delta^2)`

Also;

`\delta = √(h^2 +a^2 +2ah sin \ \theta) - √(h^2 +a^2)`

Thus;

`k = (mg(a+b) sin \ \theta )/(( √(h^2 +a^2 +2ah sin \ \theta) - √(h^2 +a^2))^2)`

where;

`\delta` = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

`k = ((1.53*9.8)(0.6+0.2) sin \ 64 )/(( √(0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64) - √(0.6^2 +0.6^2))^2)`

`k = 104.82\ \ N/m`

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

`T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = (1)/(2)I_o \omega_3^2+(1)/(2)k \delta^2 - (mg(a+b)sin \theta )/(2) \\ \\ (1)/(2) (m(a+b)^2)/(3) \omega_3^2 + (1)/(2) k \delta^2 - (mg(a+b)sin \ \theta )/(2) =0`

`(m(a+b)^2)/(3) \omega_3^2 + k(√(h^2+a^2+2ah sin \theta ) - √(h^2+a^2))^2 - mg(a+b)sin \theta = 0`

`(1.53(0.6+0.6)^2)/(3) \omega_3^2 + 104.82(√(0.6^2+0.6^2+2(0.6*0.6) sin 32 ) - √(0.6^2+0.6^2))^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0`

`0.7344 \omega_3^2 = 2.128`

`\omega _3 = \sqrt{(2.128)/(0.7344) }`

`\omega _3 =1.70 \ rad/s`

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s