a. τ = τ 2 + τ 3 − τ 1 \tau=\tau_2+\tau_3-\tau_1 τ = τ 2 + τ 3 − τ 1
τ 1 = r 1 F 1 sin α 1 \tau_1=r_1F_1\sin\alpha_1 τ 1 = r 1 F 1 sin α 1
τ 2 = r 2 F 2 sin α 2 \tau_2=r_2F_2\sin\alpha_2 τ 2 = r 2 F 2 sin α 2
τ 3 = r 3 F 3 sin α 3 \tau_3=r_3F_3\sin\alpha_3 τ 3 = r 3 F 3 sin α 3
r 1 = r 2 = r 3 = r r_1=r_2=r_3=r r 1 = r 2 = r 3 = r
r = ( L / 2 ) 2 + ( L / 2 ) 2 = L / 2 = 0.18 / 2 = 0.127 ( m ) r=\sqrt{(L/2)^2+(L/2)^2}=L/\sqrt2=0.18/\sqrt2=0.127\ (m) r = ( L /2 ) 2 + ( L /2 ) 2 = L / 2 = 0.18/ 2 = 0.127 ( m )
τ 1 = 0.127 ⋅ 18 ⋅ sin 135 ° = 1.62 ( N ⋅ m ) \tau_1=0.127\cdot 18\cdot\sin135°=1.62\ (N\cdot m) τ 1 = 0.127 ⋅ 18 ⋅ sin 135° = 1.62 ( N ⋅ m )
τ 2 = 0.127 ⋅ 26 ⋅ sin 135 ° = 2.33 ( N ⋅ m ) \tau_2=0.127\cdot 26\cdot\sin135°=2.33\ (N\cdot m) τ 2 = 0.127 ⋅ 26 ⋅ sin 135° = 2.33 ( N ⋅ m )
τ 3 = 0.127 ⋅ 14 ⋅ sin 90 ° = 1.78 ( N ⋅ m ) \tau_3=0.127\cdot 14\cdot\sin90°=1.78\ (N\cdot m) τ 3 = 0.127 ⋅ 14 ⋅ sin 90° = 1.78 ( N ⋅ m )
τ = 2.33 + 1.78 − 1.62 = 2.49 ( N ⋅ m ) \tau=2.33+1.78-1.62=2.49\ (N\cdot m) τ = 2.33 + 1.78 − 1.62 = 2.49 ( N ⋅ m ) . Answer
b. τ = I ϵ → ϵ = τ / I \tau=I\epsilon\to \epsilon=\tau/I τ = I ϵ → ϵ = τ / I
I = 1 12 m ( a 2 + b 2 ) = 1 12 ⋅ 1.2 ⋅ ( 0.1 8 2 + 0.1 8 2 ) = 0.00648 ( k g ⋅ m 2 ) I=\frac{1}{12}m(a^2+b^2)=\frac{1}{12}\cdot1.2\cdot(0.18^2+0.18^2)=0.00648\ (kg\cdot m^2) I = 12 1 m ( a 2 + b 2 ) = 12 1 ⋅ 1.2 ⋅ ( 0.1 8 2 + 0.1 8 2 ) = 0.00648 ( k g ⋅ m 2 )
ϵ = 2.49 / 0.00648 = 384 ( r a d / s 2 ) \epsilon=2.49/0.00648=384\ (rad/s^2) ϵ = 2.49/0.00648 = 384 ( r a d / s 2 ) . Answer
c. ω = ω 0 + ϵ t = 0 + 384 ⋅ 3 = 1152 ( r a d / s ) \omega=\omega_0+\epsilon t=0+384\cdot3=1152\ (rad/s) ω = ω 0 + ϵ t = 0 + 384 ⋅ 3 = 1152 ( r a d / s ) . Answer
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