Answer :
Answer:
Speed of 70g bob = 2.291m/s
The tension = 0.864 N
Explanation:
The height of rise of the pendulum is
X = L ( 1- cos θ)
X= (2)( 1 - cos30)
X = (2)(1- 8660)
X=(2)(0.134)
X=0.268
According to conservation of energy
[tex]\frac{1}{2}mV^2 = m g x[/tex]
[tex]\frac{1}{2}V^2 =\frac{ m g x}{m}[/tex]
[tex]\frac{1}{2}V^2 = g x[/tex]
[tex]V^2 = 2g x[/tex]
[tex]V = \sqrt{ 2g x}[/tex]
Substituting the values,
V =[tex]\sqrt{ 2 (9.8)(0.268m)}[/tex]
V =[tex]\sqrt{5.2528}[/tex]
V =2.291m/s
Now Tension can be calculated as
T = ma
where
m is the mass
a is the centripetal acceleration
So now ,
[tex]a= \frac{V^2}{r}[/tex]
[tex]a= \frac{2.291^2}{2}[/tex]
[tex]a=\frac{5.24}{2}[/tex]
[tex]a=2.62 m/s^2[/tex]
substituting in Tension formula,
T= (0.07)(9.8 +2.62 ) ( 70 g is 0.07 kg)
T= (0.07)(12.42)
T=0.864 N