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Artyom0805 [142]

4 years ago

5

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a

hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
a. Is angular momentum conserved?
b. Find the change in kinetic energy of the block, in J.
c. How much work was done in pulling the cord? in J.

1 answer:

sweet [91]4 years ago

8 0

Answer:

W= $K_2-K_1==9.12\times10^{-3} J$

Explanation:

a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.

b) By the law of conservation of energy angular momentum

$L_1=L_2$

$I_1\omega_1=I_2\omega_2$

$mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1$

$\omega_2=(\frac{0.3}{0.15})^2\times2.85$

$\omega_2=5.7\text{ rad/sec}$

c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)

$K_1=\frac{1}{2} mr_1^2\omega_1^2$

$K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2$

$=9.12\times10^{-3} J$

Now,

$K_2=\frac{1}{2} mr_2^2\omega_2^2$

$K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2$

$K_2=18.24\times10^{-3}$ J

Therefore, Work done W= $K_2-K_1==9.12\times10^{-3} J$

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Answer:

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Explanation:

From the law of conservation of linear momentum

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Substituting 6g=0.006 Kg for mass of bullet, 0.1Kg for mass of stone, 350 m/s for initial velocity of bullet and 250 m/s as final velocity for stone but initial velocity of stone is zero

$0.006 Kg *350 m/s (\hat i)+0.1 Kg*0=0.006 Kg* 250 m/s (\hat j)+0.1*v_{fs}$

$2.1 Kg.m/s(\hat i)=1.5 Kg.m/s(\hat j)+ 0.1*v_{fs}$

$0.1*v_{fs}=2.1 Kg.m/s(\hat i)- 1.5 Kg.m/s(\hat j)$

$v_{fs}=21 Kg.m/s(\hat i)- 15 Kg.m/s(\hat j)$

$|v_{fs}|=\sqrt {(v_{x}^{2}+v_{y}^{2})}$

Substituting 21 for $v_{x}$ and 15 for $v_{y}$

$|v_{fs}|=\sqrt {(21^{2}+15^{2})}=25.8 m/s$

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$tan\theta=\frac {v_{y}}{v_{x}}$

$\theta=tan^{-1}(\frac {15}{21})=35.5^{o}$

Therefore, magnitude is 25.8 m/s and direction is $35.5^{o}$

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Read 2 more answers

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

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Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

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