m₁a=m₁g-T₁
m₂a=T₂- F(fr)
0=m₂g-N
Iε=(T₁-T₂)R =>T₁-T₂ = Iε/R=Ia/R²
N= m₂g,
F(fr)=μN=μm₂g
m₁a+ m₂a= m₁g-T₁+ T₂- F(fr)=
=m₁g- μm₂g –( T₁-T₂) =
=m₁g- μm₂g – Ia/R².
a=g(m₁-μm₂)/{m₁+ m₂+(I/R²)}.
d=at²/2
A pulley of mass mp , radius R , and moment of inertia about its center of mass Ic , is attached to the edge of a table. An inextensible string of negligible mass is wrapped around the pulley and attached on one end to block 1 that hangs over the edge of the table. The other end of the string is attached to block 2 which slides along a table. The coefficient of sliding friction between the table and the block 2 is μk . Block 1 has mass m1 and block 2 has mass m2 , with m1>μkm2 . At time t=0 , the blocks are released from rest. At time t=t1 , block 1 hits the ground. Let g denote the gravitational acceleration near the surface of the earth.
(a) Find the magnitude of the linear acceleration of the blocks. Express your answer in terms of m1 , m2, Ic, R, μk and g as needed (enter m_1 for m1, m_2 for m2, I_c for Ic, R for R, mu_k for μk and g for g).
a=
(b) How far did the block 1 fall before hitting the ground? Express your answer in terms of m1 , m2, Ic, R, μk, t1 and g as needed (enter m_1 for m1, m_2 for m2, I_c for Ic, R for R, mu_k for μk, t_1 for t1 and g for g).
d=
1 answer
1 answer