
A phonograph record on turntable rotates 30 rpm. The linear speed of a point on the record at the needle at the beginning of the recording when it is at a distance of 14m from the centre is
A) 22cm/sec
B) 44cm/sec
C) 48cm/sec
D) 52cm/sec
Answer
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Hint: The frequency (rpm) of the recorder is given, it can be used to calculate the angular speed (as the record is rotating, it will have rotational speed called angular speed). Then using the relationship between the linear and angular speed, we can find the value of linear speed of phonograph record.
Formula Used: $\omega = 2\pi \nu $ where,
$\omega $ = Angular speed
$\pi = \dfrac{{22}}{7}$
$\nu $ = frequency (in rps)
Relationship between linear and angular speed:
$v = \omega r$ where,
v = linear speed
r = distance from the center.
Complete step by step answer:
As the record is rotating, it will have rotational speed called angular speed which is equal to:
$\omega = 2\pi \nu $ ______ (1)
Frequency $\left( \nu \right)$ :
30 rotations per minute, per second it can be calculated using unitary method
1 min = 30 rotations
or
60 seconds = 30 rotations
1 second = $\dfrac{{30}}{{60}} \times 1$
= $\dfrac{1}{2}$
The phonograph take$\dfrac{1}{2}$rotations in 1 second and thus the frequency is $\dfrac{1}{2}$
Substituting this in (1)
$\omega = 2 \times \pi \times \dfrac{1}{2}$
$\omega = \pi $
The angular speed of the recorder is $\pi $ rad/s.
Now, the relationship between linear and angular speed is:
Linear speed (v) = Angular speed $\left( \omega \right)$X radius (r) [distance from the centre]
r = 14 cm (given)
$v = \omega r$
Substituting the values, we get:
$
v = \pi \times 14 \\
v = \dfrac{{22}}{7} \times 14 \\
$
v = 44
Therefore, the linear speed of the phonograph record is 44 cm/s
Note:In unitary method, if quantity A has value x, then quantity B in terms of x can be written as:
A = x
B = $\dfrac{x}{A} \times B$
Frequency is basically the number of times a particular event occurs or repeats.
The units of angular speed $\left( \omega \right)$ is rad/s because as it is in rotational motion, the distance is covers is in radians while in the linear speed (v), the distance is covered in a line in units like m/cm/mm etc.
Formula Used: $\omega = 2\pi \nu $ where,
$\omega $ = Angular speed
$\pi = \dfrac{{22}}{7}$
$\nu $ = frequency (in rps)
Relationship between linear and angular speed:
$v = \omega r$ where,
v = linear speed
r = distance from the center.
Complete step by step answer:
As the record is rotating, it will have rotational speed called angular speed which is equal to:
$\omega = 2\pi \nu $ ______ (1)
Frequency $\left( \nu \right)$ :
30 rotations per minute, per second it can be calculated using unitary method
1 min = 30 rotations
or
60 seconds = 30 rotations
1 second = $\dfrac{{30}}{{60}} \times 1$
= $\dfrac{1}{2}$
The phonograph take$\dfrac{1}{2}$rotations in 1 second and thus the frequency is $\dfrac{1}{2}$
Substituting this in (1)
$\omega = 2 \times \pi \times \dfrac{1}{2}$
$\omega = \pi $
The angular speed of the recorder is $\pi $ rad/s.
Now, the relationship between linear and angular speed is:
Linear speed (v) = Angular speed $\left( \omega \right)$X radius (r) [distance from the centre]
r = 14 cm (given)
$v = \omega r$
Substituting the values, we get:
$
v = \pi \times 14 \\
v = \dfrac{{22}}{7} \times 14 \\
$
v = 44
Therefore, the linear speed of the phonograph record is 44 cm/s
Note:In unitary method, if quantity A has value x, then quantity B in terms of x can be written as:
A = x
B = $\dfrac{x}{A} \times B$
Frequency is basically the number of times a particular event occurs or repeats.
The units of angular speed $\left( \omega \right)$ is rad/s because as it is in rotational motion, the distance is covers is in radians while in the linear speed (v), the distance is covered in a line in units like m/cm/mm etc.
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