A tire 0.500 m in radius at a constant rate of 200 rev/min.Find the speed and acceleration of a small stone lodged in the tread of the tire( on its outer edge).
I assume you want the speed relative to the axle. The velocity with respect to the ground depends upon the angular location of the stone.
w = 20.94 radians/s
V = R*w = 10.47 m/s
a = R*w^2 = 219 m/s^2
Thank's,i do appreciate ur help.
To find the speed and acceleration of the small stone lodged in the tread of the tire, we can follow these steps:
Step 1: Calculate the circumference of the tire.
The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle. In this case, the radius of the tire is 0.500 m. Therefore, the circumference of the tire is:
C = 2π(0.500) = 3.1416 m
Step 2: Calculate the distance traveled by the small stone in one revolution of the tire.
Since the stone is lodged in the tread of the tire, it will also travel along the circumference of the tire in one revolution. Therefore, the distance traveled by the stone in one revolution is equal to the circumference of the tire.
Distance = Circumference = 3.1416 m
Step 3: Calculate the time taken for one revolution.
Since the tire is rotating at a constant rate of 200 revolutions per minute, we can find the time taken for one revolution by dividing the total time taken for one minute (60 seconds) by 200.
Time taken for one revolution = (1 minute / 200 revolutions) = 0.3 seconds
Step 4: Calculate the speed of the stone.
Speed is defined as the distance traveled divided by the time taken. In this case, the distance traveled by the stone in one revolution is equal to the circumference of the tire, which is 3.1416 m, and the time taken for one revolution is 0.3 seconds.
Speed = Distance / Time = 3.1416 m / 0.3 seconds ≈ 10.47 m/s
Step 5: Calculate the acceleration of the stone.
Acceleration can be calculated using the formula a = v^2 / r, where v is the speed and r is the radius of the circle. In this case, the speed of the stone is 10.47 m/s, and the radius of the tire is 0.500 m.
Acceleration = (10.47 m/s)^2 / 0.500 m ≈ 218.02 m/s^2
Therefore, the speed of the stone lodged in the tread of the tire is approximately 10.47 m/s, and its acceleration is approximately 218.02 m/s^2.
To find the speed and acceleration of the small stone lodged in the tread of the tire, we can use the concepts of linear speed and centripetal acceleration.
1. Finding the linear speed:
Linear speed is the distance traveled per unit time along a circular path. Since the stone is lodged on the outer edge of the tire, it moves along a circular path as the tire rotates.
The linear speed of the stone can be calculated using the following formula:
Linear speed = (radius of the tire) × (angular speed)
Given:
Radius of the tire (r) = 0.500 m
Angular speed (ω) = 200 rev/min
To convert the angular speed from revolutions per minute to radians per second, we need to use the conversion factor:
1 revolution = 2π radians
So, the angular speed in radians per second (ω_rad/s) can be calculated as follows:
ω_rad/s = (angular speed in rev/min) × (2π rad/1 rev) × (1 min/60 s)
Now we can calculate the linear speed (v):
v = r × ω_rad/s
First, calculate ω_rad/s:
ω_rad/s = (200 rev/min) × (2π rad/1 rev) × (1 min/60 s)
= (200 × 2π) rad/s
= 400π rad/s
Now, calculate the linear speed (v):
v = (0.500 m) × (400π rad/s)
≈ 628.32 m/s
Therefore, the linear speed of the small stone lodged in the tread of the tire is approximately 628.32 m/s.
2. Finding the acceleration:
Centripetal acceleration is the acceleration in the direction of the center of the circular path, caused by the change in direction of a particle moving along that path. The centripetal acceleration can be determined using the formula:
Centripetal acceleration (a_c) = (linear speed)^2 / (radius of the circular path)
The radius of the circular path in this case is the same as the radius of the tire.
Now we can calculate the centripetal acceleration (a_c):
a_c = (v)^2 / r
Substituting the values of v and r:
a_c = (628.32 m/s)^2 / (0.500 m)
≈ 786665.28 m/s^2
Therefore, the acceleration of the small stone lodged in the tread of the tire is approximately 786665.28 m/s^2.