The radius of curvature of a loop-the-loop for a roller coaster is 11.8 m. At the top of the loop (with the car inside the loop), the force that the seat exerts on a passenger of mass m is 0.80mg. Find the speed of the roller coaster at the top of the loop.
1 m/s
To find the speed of the roller coaster at the top of the loop, we can use the concept of centripetal force.
Step 1: Determine the centripetal force:
At the top of the loop, the centripetal force is provided by the net force acting on the passenger. In this case, the net force is the difference between the gravitational force and the force exerted by the seat on the passenger.
Given:
Radius of curvature (r) = 11.8 m
Force exerted by the seat (F) = 0.80mg, where g is the acceleration due to gravity (9.8 m/s^2).
The gravitational force acting on the passenger can be calculated as:
Gravitational force (mg) = mass (m) x acceleration due to gravity (g).
Step 2: Calculate the gravitational force acting on the passenger:
Gravitational force (mg) = m x g
Step 3: Calculate the net force acting on the passenger:
Net force = Gravitational force - Force exerted by the seat
Net force (F_net) = mg - F
Step 4: Calculate the speed of the roller coaster at the top of the loop using the centripetal force equation:
Centripetal force = mass (m) x velocity (v)^2 / radius of curvature (r)
F_net = m x v^2 / r
Rearranging the equation to solve for the velocity (v), we get:
v^2 = F_net x r / m
Taking the square root of both sides, we find:
v = √( F_net x r / m )
Step 5: Substitute the given values into the equation:
v = √( 0.80mg x 11.8 / m )
The mass cancels out, and we are left with:
v = √( 0.80 x 9.8 x 11.8 )
Solving this equation, we find:
v ≈ 9.76 m/s
Therefore, the speed of the roller coaster at the top of the loop is approximately 9.76 m/s.
To find the speed of the roller coaster at the top of the loop, we can use the concept of centripetal force.
At the top of the loop-the-loop, the force of gravity acting on the passenger and the normal force from the seat combine to provide the centripetal force required for circular motion. The net force acting towards the center of the loop is given by:
F_net = F_gravity + F_seat
where F_gravity = mg (force of gravity) and F_seat = 0.80mg (force from the seat).
Since the net force is responsible for the circular motion, it can be equated to the necessary centripetal force:
F_net = mv^2 / r
where m is the mass of the passenger, v is the speed of the roller coaster, and r is the radius of curvature of the loop-the-loop.
Now, we can equate the expressions for the net force and centripetal force:
mg + 0.8mg = mv^2 / r
Substituting the given values:
m = mass of the passenger
g = acceleration due to gravity (approximately 9.8 m/s^2)
F_gravity = mg
F_seat = 0.80mg
r = 11.8 m
(1 + 0.8)mg = mv^2 / 11.8
Simplifying the equation:
(1.8)mg = mv^2 / 11.8
Cross-multiplying:
1.8mg * 11.8 = mv^2
Rearranging the equation to solve for v:
v^2 = (1.8mg * 11.8) / m
v^2 = 21.24g
Taking the square root of both sides:
v = √(21.24g)
Substituting the value of g (approximately 9.8 m/s^2):
v = √(21.24 * 9.8)
Calculating the value:
v ≈ √208.152
v ≈ 14.41 m/s
Therefore, the speed of the roller coaster at the top of the loop is approximately 14.41 m/s.