For m1: m1•a=T1
For m2: m2•a=m2•g – T2
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²)
T1= m1•m2•g/(m1+m2+ I/R²)
T2 = m2•g -m2•a=...
A 9.00 kg mass is connected by a light cord to a 1.00 kg mass on a smooth surface as shown in the figure. The pulley rotates about a frictionless axle and has a moment of inertia of 0.300 k∙m2 and a radius of 0.500 m. Assuming that the cord does not slip on the pulley, find (a) the acceleration of the two masses (b) the tension T1 (attached to the hanging mass).(b) the tension T2
5 answers
I am geting .875 for my acceleration and its saying its wrong! Was your value this small?
For m1: m1•a=T1 (block m1=1 kg is on the surface )
For m2: m2•a=m2•g – T2 (m2=9 kg)
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²) =9•9.8/{1+9+(0.3/0.25)}=7.885 m/s²
T1= m1•m2•g/(m1+m2+ I/R²)
For m2: m2•a=m2•g – T2 (m2=9 kg)
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²) =9•9.8/{1+9+(0.3/0.25)}=7.885 m/s²
T1= m1•m2•g/(m1+m2+ I/R²)
FOr my Tension 1 I am getting 7.875 and its wrong, im so confused what im doing wrong!
T1=m1•a=1•7.885 = 7.885 N
0 answers