Ask question

Ask question

All categories

4 years ago

12

A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.

020 kg  m2. A string is wound around the cylinder and pulled with a force of 1.0 N. The angular acceleration of the cylinder is:

1 answer:

Bumek [7]4 years ago

5 0

Answer: $5 rad/s^2$

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

$Torque =I\alpha =F\cdot r$

where $\alpha =angular\ acceleration$

$I\alpha =F\cdot r$

$0.02\cdot \alpha =1\cdot 0.1$

$\alpha =5 rad/s^2$

You might be interested in

A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.

puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

5 0

3 years ago

What is the primary outgoing radiation put off by the sun?

ira [324]

I really don’t know but I think it’s D

7 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

2 years ago

En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

5 0

3 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

4 years ago