arthham

JEE Advanced 2020 · System of Particles and Rotational Motion

Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm0.00\,\text{cm} and the right one at 90.00 cm90.00\,\text{cm}. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance xRx_R from the center (50.00 cm50.00\,\text{cm}) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.400.40 and 0.320.32, respectively, the value of xRx_R (in cm) is ______.

Answer: 25.625.6

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

Step 1: Define the initial coordinates and parameters.

The uniform meter scale has its center of mass at 50.00 cm50.00\,\text{cm}.

y0=50.00 cmy_0 = 50.00\,\text{cm} z0=40.00 cmz_0 = 40.00\,\text{cm} μs=0.40\mu_s = 0.40 μk=0.32\mu_k = 0.32

Step 2: Determine the normal forces on each finger.

Taking torque about the contact points of the left and right fingers yields the normal forces.

NL=Mgzy+zN_L = \frac{Mgz}{y+z} NR=Mgyy+zN_R = \frac{Mgy}{y+z}

Step 3: Establish the condition for slipping.

The left finger slips while the right finger grips as long as the kinetic friction on the left is less than or equal to the maximum static friction on the right.

μkNL≤μsNR\mu_k N_L \le \mu_s N_R μkz≤μsy\mu_k z \le \mu_s y

Step 4: Calculate the position where the left finger stops slipping.

The left finger slips, decreasing yy from y0y_0 while zz remains constant at z0z_0, until the slipping condition reaches its limit.

μkz0=μsy1\mu_k z_0 = \mu_s y_1 0.32×40=0.40×y10.32 \times 40 = 0.40 \times y_1 y1=32 cmy_1 = 32\,\text{cm}

Step 5: Calculate the position where the right finger stops slipping.

The right finger now slips, decreasing zz from z0z_0 while the left finger remains stationary at y1y_1, until the roles reverse again.

μky1=μsz1\mu_k y_1 = \mu_s z_1 0.32×32=0.40×z10.32 \times 32 = 0.40 \times z_1 z1=25.6 cmz_1 = 25.6\,\text{cm}

The distance of the right finger from the center at this instant is xR=z1x_R = z_1.

xR=25.6 cmx_R = 25.6\,\text{cm}

Answer: 25.6

More from System of Particles and Rotational Motion

All 46 System of Particles and Rotational Motion questions →