A block slides down a frictionless ramp. It starts from the top, at rest. It is

Question

A block slides down a frictionless ramp. It starts from the top, at rest. It is connected to a pulley by a rope that wraps many times around the pulley and unwinds as the mass slides down. The length L = 0.8m and the angle ? is 30 degrees. The pulley is a solid cylinder of radius 2.0cm and mass m=0.15 kg. The block has mass M=0.55 kg. The pulley turns without friction and the rope doesn't slip; the mass slides without friction. A) What is the speed of the block when it reaches the bottom? For this part, use force and torque methods. B) Using an energy method instead, calculate the same answer

Explanation / Answer

A)

Let the speed of the block when it reaches bottom be v.

Net force on the block along the slope,

Fnet = Mg*sin(30) - T = M*a

where T = tension in the string

Also, for pulley, torque = T*R = I*A

where I = moment of inertia of pulley = 0.5*mR^2

A = angular acceleration,

for pulley, A = a/R

where a = acceleration of block

So, T*R = (0.5*mR^2)*(a/R) = 0.5*maR

So, T = 0.5*ma

So, Mg*sin(30) - 0.5*ma = Ma

So, a = Mg*sin(30)/(0.5*m+M)

Using equation of motion,

v^2 = u^2 + 2as

where u = 0

s = L

So, v^2 = 2*(a)*L

So, v^2 = 2*( Mg*sin(30)/(0.5*m+M))*L

So, v^2 = 2*(0.55*9.8*sin(30 deg)/(0.5*0.15+0.55))*0.8 = 6.9

so, v = 2.63 m/s <---------answer

B)

We will use conservation of energy to solve this question.

Initial Kinetic energy of the block and pulley system, KEi = 0 <----- block starts from rest

Initial Potential energy of block, PEi = M*g*L*sin(30) <------- height of the slope = L*sin(30)

So, total energy of the block initially, TEi = KEi + PEi = MgL*sin(30)

Now after reaching the bottom,

Final Kinetic energy of the block = 0.5*M*v^2

and Kinetic energy of the pulley = 0.5*(I)*W^2

for a pulley, moment of inertia, I = 0.5*m*R^2

and angular speed, W = v/R

So, Kinetic energy of pulley, = 0.5*(0.5*mR^2)*(v/R)^2 = 0.25*mv^2

So, total final kinetic energy of the system(pulley + block) = 0.5*Mv^2 + 0.25*mv^2

Final potential energy = 0

So, final total energy, TEf = 0.5*Mv^2 + 0.25*mv^2

By conservation of energy, total energy remains conserved,

So, TEf = TEi

So,  0.5*Mv^2 + 0.25*mv^2 = MgL*sin(30)

So, v^2 = MgL*sin(30)/(0.5*M + 0.25*m)

So, v^2 = 0.55*9.8*0.8*sin(30 deg)/(0.5*0.55+0.25*0.15) = 6.9

So, v = 2.63 m/s <--------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.