Solving Physics Problems: Problem 3 & 5

[mustang]

Problem 3.
A dog sits 1.4 m from the center of a merry-go-round. The dog undergoes a 1.7 m/s^2 centripetal acceleration. What is the angular speed of the merry-go-round? Answer in units of rad/s.

Problem 5.
A girl sits on a tire that is attached to an overhanging tree limb by a rope 2.30 m in length. The girl's father pushes her with a tangential speed of 3.05 m/s. Besides the force opposing the girl's weight, the magnitude of the force that maintains her circular motion is 84.7 N.
What is the girl's mass? Answer in units of kg.
Note: How would you solve this?

 

[ahrkron]

In order to get any help, you need to show some progress. What have you tried?

 

[M98Ranger]

For Problem 3, we can use the formula for centripetal acceleration, a = ω^2r, where ω is the angular speed and r is the distance from the center. Plugging in the given values, we get ω = √(a/r) = √(1.7 m/s^2 / 1.4 m) = 1.15 rad/s.

For Problem 5, we can use the formula for tangential acceleration, a = αr, where α is the angular acceleration and r is the distance from the center. We can also use the formula for centripetal force, F = mv^2/r, where m is the mass and v is the tangential speed. Since the force opposing the girl's weight is not mentioned, we can assume that it is negligible and only consider the centripetal force. Setting these two formulas equal to each other, we get αr = mv^2/r. Rearranging, we get m = αr^2/v^2. Plugging in the given values, we get m = (84.7 N * 2.30 m) / (3.05 m/s)^2 = 12.7 kg. This is the mass of the girl.