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$$
\begin{array}{l}{\text { 1.3 - } 6.3 \text { . The pendulum bob } B \text { has a weight of } 5 \text { lb and is }} \\ {\text { released from rest in the position shown, } \theta=0^{\circ} . \text { Determine }} \\ {\text { the tension in string } B C \text { just after the bob is released, }} \\ {\theta=0^{\circ}, \text { and also at the instant the bob reaches } \theta=45^{\circ} \text { . }} \\ {\text { Take } r=3 \mathrm{ft} \text { . }}\end{array}
$$

$$
w=5lb
$$

$$
\theta_{0}=0^{\circ}
$$

$$
B C
$$

$$
\theta=45^{\circ}
$$

$$
r=3 f t
$$

$$
v _0=0
$$

\( \Sigma F_{n}=m a_{n} \Rightarrow ∴ T :\left(\frac{5}{32.2}\right)\left(\frac{0^{2}}{3}\right)=0 \)

$$
\Sigma F_{t}=m a _t \Rightarrow 5 \cos \theta=\left(\frac{5}{32.2}\right) a t \Rightarrow {a t}=32.2 \cos \theta
$$

$$
\Sigma  F_{n}=m a_{n} \Rightarrow T-5 \sin \theta=\left(\frac{5}{32.2}\right)\left(\frac{v^{2}}{3}\right) \Rightarrow (1)
$$

\(∵ vdv=at \ ds \Rightarrow ∵ ds=3d \theta\Rightarrow ∴ v d v=3 a+d \theta \)

\(∴ \int_{0}^{v} v d v=3(32.2) \int_{0}^{45} \cos \theta d \theta \)

\(∴ \frac{v^{2}}{2}=3(32.2)[\sin 45-\sin 0] \Rightarrow ∴ {v}=136.61 \mathrm{ft}^{2} / \mathrm{s}^{2} \)

\(∴ T-5 \sin 45=\frac{5}{32.2}\left(\frac{136.61}{3}\right) \)

$$
T=10.6l b
$$

13-87. The path of motion of a $$5-lb$$ particle in the horizontal plane is described in terms of polar coordinates as $$r=(2t+1)ft$$ and $$\theta=(0.5t^{2}-t) rad$$. where $$t$$ is in seconds. Determine the magnitude of the unbalanced force acting on the particle when $$t=2 s.$$

$$
w=5l b
$$

$$
r=(2 t+1) ft \quad an d
$$

$$
\theta=\left(0.5 t^{2}- t\right) {rad}
$$

$$
@ t=2s
$$

\( {r=2 T+1 \quad ∴ @ t=2 S \Rightarrow r=2 \times 2+1=5 \mathrm{ft}} \)

\(∴ \dot r=2 f t \quad, r^{..}=0 \)

\( \theta=0.5 t^{2}-t ∴ @ t=2s \Rightarrow 0 \)

\(∴ \dot \theta=t-\left.1\right|_{t=2}=1rad/s \Rightarrow \theta^{..}=1rad/s^2 \)

$$
a_{r}=r^{..}-r \dot \theta^{2}=0-5(1)^{2}=-5 f t / s^{2}
$$

$$
{a_{\theta}=r \theta^{..}+2 \dot r \dot \theta=5(1)+2(2)(1)=9ft/ s^{2}}
$$

$$
\Sigma F_{r}=m a_{r} \Rightarrow F_{r}=\frac{5}{32 \cdot 2}(-5)=-0.7764 \mathrm{lb}
$$

\( \Sigma F_{\theta}=ma_ \theta \Rightarrow ∴ F_{\theta}=\frac{5}{32 \cdot 2}(9)=1.398l b \)

\(∴ F=\sqrt{F_{r}^{2}+F_{\theta}^{2}}=1.6 lb \)

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