【Physics】1. Two Blocks and a Pulley、Sliding Off a Sphere

⚛️ Physics

Problem 1 (Easy–Medium): Two Blocks and a Pulley

Two blocks of massesm1=3 kgm_1 = 3\,\text{kg}m1​=3kgandm2=5 kgm_2 = 5\,\text{kg}m2​=5kgare connected by a light inextensible string over a frictionless pulley (an Atwood machine). The system is released from rest.

(a) What is the acceleration of the system?

(b) What is the tension in the string?

Hint:Write Newton’s second law for each block separately, then eliminate the tension (or the acceleration) to solve.

Solution

m2g−T=m2a,T−m1g=m1am_2g-T=m_2a, T-m_1g=m_1am2​g−T=m2​a,T−m1​g=m1​asoa=m2−m1m1+m2ga=\frac{m_2-m_1}{m_1+m_2}ga=m1​+m2​m2​−m1​​g, form1m_1m1​the acceleration is upward, form2m_2m2​downward.T=2m1m2m1+m2gT=\frac{2m_1m_2}{m_1+m_2}gT=m1​+m2​2m1​m2​​g.

Problem 2 (Medium–Hard): Sliding Off a Sphere

A small block of massmmmis placed at the very top of a frictionless sphere of radiusRRRthat is fixed to the ground. The block is given a tiny nudge and begins to slide.

At what angleθ\thetaθ(measured from the vertical) does the block lose contact with the sphere?

Hint:At the point of detachment the normal force is zero. Use energy conservation to express velocity as a function ofθ\thetaθ, then apply the radial (centripetal) direction of Newton’s second law.

Solution

Centripetal acceleration:gcos⁡θ=v2Rg\cos \theta=\frac{v^2}{R}gcosθ=Rv2​, conservation of kinetic energy:mgR(1−cos⁡θ)=12mv2mgR(1-\cos\theta)=\frac{1}{2}mv^2mgR(1−cosθ)=21​mv2,cos⁡θ=23\cos \theta=\frac{2}{3}cosθ=32​,θ=arccos⁡23\theta =\arccos\frac{2}{3}θ=arccos32​.