Answer :
Certainly! Let's solve the problem step by step.
1. Understanding the Problem:
We need to find the velocity of a point on the edge of a rotating platter. We're given:
- The period of rotation [tex]\( T = 3.86 \)[/tex] seconds. The period is the time it takes to make one complete revolution.
- The radius of the platter [tex]\( r = 15.1 \)[/tex] cm.
2. Calculate the Circumference:
The circumference of the platter (the path a point on the edge travels in one complete revolution) is given by the formula for the circumference of a circle:
[tex]\[ \text{Circumference} = 2 \pi r \][/tex]
Plugging in the radius:
[tex]\[ \text{Circumference} = 2 \pi \times 15.1 \, \text{cm} \][/tex]
3. Determine the Velocity:
The velocity (or linear speed) of a point on the edge of the platter is the distance it travels (the circumference) divided by the time it takes (the period):
[tex]\[ v = \frac{\text{Circumference}}{\text{Period}} \][/tex]
Substituting the values we have:
[tex]\[ v = \frac{2 \pi \times 15.1 \, \text{cm}}{3.86 \, \text{s}} \][/tex]
We already have the numerical results from the calculations:
- The circumference is approximately [tex]\( 94.876 \, \text{cm} \)[/tex].
- The velocity is approximately [tex]\( 24.579 \, \text{cm/s} \)[/tex].
Therefore, the velocity of a point on the edge of the platter is:
[tex]\[ v \approx 24.579 \, \text{cm/s} \][/tex]
1. Understanding the Problem:
We need to find the velocity of a point on the edge of a rotating platter. We're given:
- The period of rotation [tex]\( T = 3.86 \)[/tex] seconds. The period is the time it takes to make one complete revolution.
- The radius of the platter [tex]\( r = 15.1 \)[/tex] cm.
2. Calculate the Circumference:
The circumference of the platter (the path a point on the edge travels in one complete revolution) is given by the formula for the circumference of a circle:
[tex]\[ \text{Circumference} = 2 \pi r \][/tex]
Plugging in the radius:
[tex]\[ \text{Circumference} = 2 \pi \times 15.1 \, \text{cm} \][/tex]
3. Determine the Velocity:
The velocity (or linear speed) of a point on the edge of the platter is the distance it travels (the circumference) divided by the time it takes (the period):
[tex]\[ v = \frac{\text{Circumference}}{\text{Period}} \][/tex]
Substituting the values we have:
[tex]\[ v = \frac{2 \pi \times 15.1 \, \text{cm}}{3.86 \, \text{s}} \][/tex]
We already have the numerical results from the calculations:
- The circumference is approximately [tex]\( 94.876 \, \text{cm} \)[/tex].
- The velocity is approximately [tex]\( 24.579 \, \text{cm/s} \)[/tex].
Therefore, the velocity of a point on the edge of the platter is:
[tex]\[ v \approx 24.579 \, \text{cm/s} \][/tex]