Calculating Speedboat's Turn Dynamics

[mattmannmf]

A speedboat with a length of 10.2m and a mass of 249 kg is negotiaging a circular turn (radius = 32 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 597 N to be applied to the boat. The intial speed of the boat going into the turn is 4 m/sec.
a) What is the tangential acceleration of the boat, 2 sec into the turn?

b) What is the tangential velocity of the boat, 2 sec into the turn?

c) What is the centripetal acceleration, 2 sec into the turn?

d) What is the total acceleration, 2 sec into the turn?

for a, Tan Accel= pi*angular acceleration. but i don't know really how to get started.

 

[LowlyPion]

A speedboat with a length of 10.2m and a mass of 249 kg is negotiaging a circular turn (radius = 32 m) around a buoy. During the turn, the engine causes a net tangential force of magnitude 597 N to be applied to the boat. The intial speed of the boat going into the turn is 4 m/sec.
a) What is the tangential acceleration of the boat, 2 sec into the turn?

b) What is the tangential velocity of the boat, 2 sec into the turn?

c) What is the centripetal acceleration, 2 sec into the turn?

d) What is the total acceleration, 2 sec into the turn?

for a, Tan Accel= pi*angular acceleration. but i don't know really how to get started.

Maybe this will help?
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq

 

[nlightner]

I would approach this problem by using Newton's laws of motion and the equations for circular motion. First, I would calculate the angular acceleration of the boat using the given tangential force and the boat's mass:

Angular acceleration = net tangential force / mass
= 597 N / 249 kg
= 2.4 rad/s^2

a) To find the tangential acceleration, we can use the equation a = r * angular acceleration, where r is the radius of the turn. In this case, the tangential acceleration would be:

a = (32 m) * (2.4 rad/s^2)
= 76.8 m/s^2

b) The tangential velocity of the boat can be found using the equation v = r * angular velocity. Since the boat has been turning for 2 seconds, we can find the angular velocity using the equation:

Angular velocity = initial angular velocity + angular acceleration * time
= 0 + (2.4 rad/s^2) * (2 sec)
= 4.8 rad/s

Therefore, the tangential velocity would be:

v = (32 m) * (4.8 rad/s)
= 153.6 m/s

c) The centripetal acceleration can be found using the equation:

Centripetal acceleration = (tangential velocity)^2 / radius
= (153.6 m/s)^2 / 32 m
= 738.3 m/s^2

d) The total acceleration can be found using the Pythagorean theorem, since the tangential and centripetal accelerations are perpendicular to each other. Therefore, the total acceleration would be:

Total acceleration = √(tangential acceleration)^2 + (centripetal acceleration)^2
= √(76.8 m/s^2)^2 + (738.3 m/s^2)^2
= 743.5 m/s^2

In summary, the tangential acceleration of the boat 2 seconds into the turn is 76.8 m/s^2, the tangential velocity is 153.6 m/s, the centripetal acceleration is 738.3 m/s^2, and the total acceleration is 743.5 m/s^2.