Yo-yo physics questions?

[匿名]

2017-02-12 4:30 am

A. Assume: Drag(friction) is negligible.
Given g = 9.81 m/s^2. The density of a large Yo-yo like solid is uniform throughout. The Yo-yo like solid has a mass of 2.1 kg. A cord is wrapped around the stem of the Yo-yo like solid and attached to the ceiling. The radius of the stem is 3m and the radius of the disk is 7m. Calculate the moment of inertia about the center of mass(axis of rotation). Answer in units of kg*m^2.

B. What is the vertical acceleration of the center of mass of the Yo-you? Answer in units of m/s^2.

C. What is the magnitude of the torque the cord exerts on the center of mass of the Yo-Yo? Answer in units of Nm

D. The Yo-Yo is released from rest at height h. Find the velocity v of the center of mass of the disk at the height ∆h = 19 m. Answer in units of m/s.

回答 (1)

NCS

2017-02-13 12:09 am

✔ 最佳答案

A) In order to reach a solution, you need to make some assumption about how the mass is allocated between the stem and the two disks. I will assume the stem is massless and that the disks weigh a combined 2.1 kg.

Then the moment of inertia of the yo-yo about its center is
Ic = ½MR² = ½ * 2.1kg * (7m)² = 51.45 kg·m² ◄

B) If we consider that the yo-yo is rotating not about its center but about the point at which the string leaves the stem, then by the parallel axis theorem
I = Ic + mr² = 51.45kg·m² + 2.1kg*(3m)² = 70.35 kg·m²

torque = I*α = T*R
where T is the tension, so
70.35kg·m² * α = T*R

But α = a/R
and T = m(g - a), so
70.35kg·m² * a/3m = 2.1kg*(9.81m/s² - a)*3m
which solves to
a = 2.08 m/s² ◄

C) T = 2.1kg * (9.81 - 2.08)m/s² = 16.2 N ◄

D) v = √(2ah) = √(2 * 2.08m/s² * 19m) = 8.89 m/s ◄

It really comes down to what assumption you make about the distribution of mass.

Hope this helps!

參考: http://www.wolframalpha.com/input/?i=70.35+*+a%2F3+%3D+2.1*(9.81+-+a)*3

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