Question Solved1 Answer As you will learin in Chapter 3 ( , the angular acceleration of a simple pendulum is given by \( \ddot{\theta}=-(g / L) \) sin \( \theta \), where \( g \) is the acceleration of gravity and \( I \) is the length of the pendulum cord. Problem \( 2.72 \) Derive the expression of the angular velocity \( \theta \) as a function of the angular coordinale \( As you will learin in Chapter 3 ( , the angular acceleration of a simple pendulum is given by \( \ddot{\theta}=-(g / L) \) sin \( \theta \), where \( g \) is the acceleration of gravity and \( I \) is the length of the pendulum cord. Problem \( 2.72 \) Derive the expression of the angular velocity \( \theta \) as a function of the angular coordinale \( \theta \). The initial conditions are \( (0)=\omega_{0} \) and \( \theta(0)=\theta_{0 .} \) Problem 2.73i Let the length of the pendulum cord be \( 1=1.5 \mathrm{~m} \). If \( \theta=3.7 \mathrm{rad} / \mathrm{s} \) when \( \theta=14^{\circ} \), determine the maximum value of \( \theta \) achieved by the pendulum. The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let \( L= \) \( 5.3 \mathrm{ft} \) and assume that the pendulum is placed in motion at \( \theta=0^{\circ} \). What is the minimum angular velocity at this position for the pendulum to swing through a full circle?

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Transcribed Image Text: As you will learin in Chapter 3 ( , the angular acceleration of a simple pendulum is given by \( \ddot{\theta}=-(g / L) \) sin \( \theta \), where \( g \) is the acceleration of gravity and \( I \) is the length of the pendulum cord. Problem \( 2.72 \) Derive the expression of the angular velocity \( \theta \) as a function of the angular coordinale \( \theta \). The initial conditions are \( (0)=\omega_{0} \) and \( \theta(0)=\theta_{0 .} \) Problem 2.73i Let the length of the pendulum cord be \( 1=1.5 \mathrm{~m} \). If \( \theta=3.7 \mathrm{rad} / \mathrm{s} \) when \( \theta=14^{\circ} \), determine the maximum value of \( \theta \) achieved by the pendulum. The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let \( L= \) \( 5.3 \mathrm{ft} \) and assume that the pendulum is placed in motion at \( \theta=0^{\circ} \). What is the minimum angular velocity at this position for the pendulum to swing through a full circle?
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Transcribed Image Text: As you will learin in Chapter 3 ( , the angular acceleration of a simple pendulum is given by \( \ddot{\theta}=-(g / L) \) sin \( \theta \), where \( g \) is the acceleration of gravity and \( I \) is the length of the pendulum cord. Problem \( 2.72 \) Derive the expression of the angular velocity \( \theta \) as a function of the angular coordinale \( \theta \). The initial conditions are \( (0)=\omega_{0} \) and \( \theta(0)=\theta_{0 .} \) Problem 2.73i Let the length of the pendulum cord be \( 1=1.5 \mathrm{~m} \). If \( \theta=3.7 \mathrm{rad} / \mathrm{s} \) when \( \theta=14^{\circ} \), determine the maximum value of \( \theta \) achieved by the pendulum. The given angular acceleration remains valid even if the pendulum cord is replaced by a massless rigid bar. For this case, let \( L= \) \( 5.3 \mathrm{ft} \) and assume that the pendulum is placed in motion at \( \theta=0^{\circ} \). What is the minimum angular velocity at this position for the pendulum to swing through a full circle?
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Qsotn:- Angulas acc of massless rigid ba-{:[" Asso, "{:[theta^(˙)=-g∣L sin theta],[(d(theta^(˙)))/(dt)=-g//L sin theta]:}],[theta^(˙)=(d theta)/(dt)],[dt=(d theta)/((theta^(˙)))],[{:[" From eqn (1) \& (1) "],[(d(theta^(˙)))/(((d theta)/(d(theta^(˙)))))=-g//L quad sin theta]:}],[=>quadtheta^(˙)dtheta^(˙)=-g//L sin d theta],[" integrating eqn (3) "],[int_(theta_(0))^(theta^(˙))theta^(˙)dtheta^(˙)=int_(theta_(0))^(theta_(0))-gl_(L)sin theta d theta],[=>(theta^(˙)^(2))/(2)int_( hat(theta)_(0)) ... See the full answer