A mass 8.5 kg is attached to a light cord, which is wound around a pulley. The p

Question

A mass 8.5 kg is attached to a light cord,
which is wound around a pulley. The pulley is
a uniform solid cylinder of radius 7.4 cm and
mass 1.7 kg.The acceleration of gravity is 9.8 m/s2 .
What is the magnitude of the net torque on
the system about the point on the axis at the
center of the pulley? Answer in units of Nm.

When the mass has a speed v = 33 m/s, the
pulley has an angular speed omega=v/R
Determine the magnitude of the total an-
gular momentum of the system about a point
on the axis at the center of the pulley?
Answer in units of kgm2/s.

Using the fact that T=dl/dt and your re-
sult from the previous section, calculate the
magnitude of the acceleration of the mass.
Answer in units of m/s2.

Explanation / Answer

Mass attached to the cord is m = 8.5 kg radius r = 7.4 cm Mass of the pulley is m_p = 1.7 kg a ) Net torque is = T * R                              = ( mg - ma ) r                              = 8.5 kg ( 9.8 m/s^2 - a ) * 7.4 *10^-2 m ......(1) b ) Moment of inertia is I = m_p *r^2 Angular momentum L = I                                       = m_p r^2 ( v / r )                                       = m_p v r                                      = 1.7 kg * 33.0 m/s * 7.4*10^-2 m                                       = 4.15 kg m^2 / s c ) the magnitude of the acceleration of the mass.            = d L / dt               = I d / dt                = I    a = mg L ( m + m_p) / g m * m_p ( m + m_p)       = 8.5 * 9.8 * 4.15 kg m^2 /s ( 8.5 + 1.7 ) / 9.8 * 8.5 * 1.7 ( 8.5 +1.7 )      = 2.44 m/s^2   To solve for part a a ) 8.5 kg ( 9.8 m/s^2 - a ) * 7.4 *10^-2 m       = 8.5kg ( 9.8 - 2.44) * 7.4*10^-2 m           = 4.62 N m

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